What This Text Covers . • •
A Review of the Derivative -------------------------------------------------------------------------------------------Here the basic rules for calculating derivatives will be reviewed_
Implicit Functions and Parametric Equations --------------,--------------------------------------------------- 2 In this section, you will Jearn to calculate the first and second derivatives of implicit
functions and the derivatives of parametric equations by use of the chain rule_
The Derivatives of Trigonometric Functions ------------------------------------------------------------------ 9 This section deals with the derivatives of the six trigonometric functions and the six in
verse trigonometric functions_
The Derivative of Exponential and Logarithmic Functions -------------------------------------------- 26 Here you will first review the meaning of exponential functions and logarithmic func
tions_ You will then Jearn to calculate the derivatives of exponential and logarithmic
functions_
Applications of the Derivative ---------------------------------------------------------------------------------------- 39 This section starts with a review of the meaning of polar coordinates, and then explains
the use of polar differentials_ As you will see, polar differentials may be used to calculate
angles to the tangent line of a curve, angular velocity and acceleration, the relations be
tween the rectangular and angular components of velocity, the acceleration of a re
volving body, and circular motion with constant angular acceleration_
The Derivatives of Hyperbolic Functions ---------------------------------------------------------------------- 59 After you study the definitions of the hyperbolic functions, you will Jearn to calculate
the derivatives of the hyperbolic functions and of their inverses_
Appendix ------------------------------------------------------------------------------------------------------------------------ 65 Here you will find the solutions to the self-tests given in the text_
CALCULUS: FUNCTION AND USE Part 3
A Review of the Derivative
1. Student Objectives
In Part 1 we began the study of the method of calculus by considering the derivatives of polynomials. In Part 2 we learned the reverse process, integration, and
many of its applications. In this text we will consider the derivatives of other func
tions, such as trigonometric, exponential, and hyperbolic functions and we will also discuss applications of these derivatives to angular measure.
When you complete this text, you will be able to:
• Calculate the derivative of an implicit function. • Use the chain rule to calculate the derivative of a function defined by para
metric equations. • Calculate the derivative of any trigonometric function, any exponential
function with base e, any natural logarithmic function, and any hyperbolic function.
• Calculate the derivative of any compounded function. • Calculate the derivative of a function that is the product of any two or more
functions. • Calculate the derivative of any implicit function. • Calculate the angle that a tangent line to a curve at a point makes with the
point's radius vector. • Calculate the angular acceleration of a line when its angular position is de
fined as a function of time.
Before moving on to derivatives of transcendental functions, we will review the formulas for calculating derivatives that we studied in Part 1 of this series.
2. Common Formulas for Derivatives
There are six essential formulas for calculating derivatives that you have learned
so far. Here the six formulas are listed in both derivative and differential form. You should have these formulas memorized by now.
du dv
de !(~)= v--u-
dx dx -=0 (1) (5) dx v2
d du d (du) - (cu)=c- (2) dx (un) = nun-1 dx (6) dx dx
d du dv d(c) = 0 -(u+v)=-+- (3) (7)
dx dx dx
d dv du d(cu) = c du - (uv) = u - + v- (4) (8)
dx dx dx
d(u+v)=du+dv (9) a(~\= v_du---;:;-_u_d_v v} v2
(11)
d(uv) = v du + u dv (10) (12)
Read through the list of formulas. Then close your text and write out each formula on a clean piece of paper. Check the formulas you wrote with the formulas listed to be sure that you have memorized them correctly.
The following self-test is designed to check your ability to apply the preceding formulas for differentiation. To get the given answer some extra algebraic manipulation may be required for some of the problems. You are urged to test your strength in algebra as well as in differentiation.
Find: in each of the following:
1. y = bx 3
2. y = b(x + 2) 3
3. y = 16(x 3 + 2x) 4
4. y = (x + bt (x 2 - cr
Self-Test 1
5. y = (3x 2 + 2x - 3)2 + .J x 2 + 12
6. y = (x2- 1)2(x2 + 2)3
(3x + 1)2
7. y= (2x2- 1)3
8. y = x(x 2 + 4x- 1)512
Implicit Functions and Parametric Equations
3. Change of Variable in Differentiation
Let us use the formula for the derivative of a product to see what the result will be when we change the independent variable. Equation 4, Art. 2, was given as
Letting u = x and v = y, then
dx since-= 1
dx
d du dv - (uv) = v- + udx dx dx
d dx dy - (xy)=y-+xdx dx dx
dy =y+x-
dx (13)
If we needed to find the derivative of the same product (x y) with respect toy, then
. dy smce- = 1
dy
d dx dy -(xy)=y-+xdy dy dy
dx =y- +x
dy
2
( 14)
and
Let us repeat the steps, but with a different product, such asx2 y 3 • Thus,
!!.cx2y3) = y3(2x) dx + x2(3y2) dy dx dx dx
= 2xy3 + 3x2 y2 dy dx
(15)
(16)
It should be both informative and interesting to note that equations 13 and 14 are
identical in differential form, which is
d(xy) = y dx + x dy (17)
A similar situation exists with equations 15 and 16:
(18)
The differential form is very useful, especially when the anti-derivative is desired, because
dx and dy, which are important parts of the integral, appear in the equation.
4. Differentiation of Implicit Functions
In most of the equations you have studied so far, the dependent variable was
expressed as an explicit function of the independent variable. Often it is either not
convenient or not possible to express one variable as an explicit function of the
other. In such situations the usual rules for finding the derivative can be applied to
each side of the equation and the desired derivative found as an implicit function
of the variables involved. Thus the equation may be differentiated as given. This
differentiation will result in another equation, which may be solved for the desired
derivative. This method can best be illustrated by the following examples:
Example 1
Given the equation of a circle x 2 + y 2 = 16, find the derivative dx· Solution
Since we are to find ;ix, we differentiate both sides of the equation with respect to x; thus
or
Therefore,
! (x2 + y2)= !(16)
dy 2x + 2y dx = 0
dy =-~ dx y
J
Ans.
Example 2
Find the second derivative,::~. if x 2 + y 2 = 16.
Solution dy X
From example 1 we know that -d = - -. Therefore, the second derivative is X y
dy X But-=·--· thus
dx y' '
Example 3
:x(d;) =~~=- !~) dy
y-xdx =-yr-
Ans.
dy dx Given the equation x3y 2 = 16. Find-;- and d-, each as a function of x andy.
ax y
Solution
Let us begin the solution by using differentials; thus,
d(x 3y 2) = d (16)
3x 2y 2 dx + 2x 3y dy = 0
Dividing by x 2y and transposing one·term, we get
Dividing both sides by 2x dx, we get
dx . dy For dy , mvert dx and get
Example 4
2xdy =- 3ydx
dy =- 3y dx 2x·
dx 2x dy =- 3y"
Ans.
Ans.
Find the equation of the tangent line to the curve x 5 - y 4 + x 3 - y 2 = 0 at the point (1, 1 ).
Solution
Using differentials on the given equation, we get Sx4 dx- 4y3 dy + 3x2 dx- 2y dy = 0
Factoring,
Dividing by dx,
- (4y 3 + 2y)dy =- (5x 4 + 3x2)dx
dy (5x 4 + 3x2) dx = (4y3 + 2y)
Given, when x = 1 andy = 1 : = *= T• the slope of the tangent line. Using the point-slope form for
the tangent line, we get
4
Simplifying, we get
3y = 4x - 1. Ans.
This is the desired equation for the tangent tine to the curve.
Self-Test 2
Find: from each of the following equations:
1. (x- y)y2 + x + y = 0 x 3 y a
4. -+-=-y3 X b
2. y2 = x + v' x2 + y2 5. (x+y)(x-y)2 =b2
3. (y + x)lf2 + (y _ x)lf2 =a 6. x 4 -4x2y 2 +y 3 =0
. dy d2y . . Fmd- and - 2 from each of the followmg equations:
dx dx
7. 2x + xy + 3y = 6
8. y =.JX+Y 9. Find the equation of the tangent line drawn to the circle whose equation isx2 - 4x
+ y 2 + 6y = 12 at the point (5, 1).
10. Find the equation of the tangent line drawn to the curve y 3 = 4x2 - yx 2 at the
point (- 2, 2).
5. Introduction to Parametric Equations
When a particle moves in a curved plane path rather than in a straight lirie, it is cus
tomary to specify its position at any time t by means of equations which give both x and
y as functions of t. For example, the equations of the trajectory of a projectile shown in
Fig. 1 have the form x = V0xt andy= V0 yt- ~ct2 , where Vox and V0 y are the x andy
components of V0 , which is the initial velocity at time t = 0.
y
X
Fig. I A trajectory with an initial velocity V 0
5
More generally, when discussing equations giving x andy as functions of time, we
write them in the general form x = g(t) andy = h(t). These two equations would specify
(x,y) in terms oft. Such equations are called parametric equations, and t is called the
parameter. Often we can eliminate t from the equations and obtain an equation of the
form y = f(x), in whichy is a function of x. From this equation we can get the derivative
dy = f'(x), which is the slope of the curve y = f(x). At this point, a question should dx ·
dy dy dx present itself concerning how dx is related to - = h'(t) and - = g'(t). The answer
dt dt
to this question is contained in the next article on the chain rule for derivatives. But be
fore getting into that, let us develop the equation y = f(x) from the two given equations . X
for the trajectory, x = V0xt andy= V0yt- 1;2ct2. If x = V0xt, then t =-,which, when Vox
substituted into the equation for y, gives
y = V0 yt- Y2ct2
gives
x x 2 y = Voy- -Y2 c-2 = j(x)
Vox Vox
and the derivative is
dy Voy x , -=- -c- =f(x) dx Vox V2 Ox
6. Chain Rule for Derivatives
We have applied the chain rule for calculating the derivatives of compound
functions. Now we shall formalize this rule so that we can use it to calculate deriva
tives of parametric functions.
RULE: If y = f(x) is a differentiable function of x. and x = g(t) is a differentiable
function oft, then y = f[g(t)] = h(t), which is also a differentiable function oft, and its
derivative with respect to time ish '(t)= f'(x)g '(t), or, in other symbols, dy = (dy) ( dx)· dt dx dt
This is called the chain rule of differentiation because it involves a chain of steps.
Ut us review these steps for finding the derivative of y with respect to t when y = f(x)
and x = g(t), or in short form, y = f[g(t) ]. We can see that here y is a function of a func
tion, which is why we need to take the following three steps to find the derivative of y
with respect to t.
First, differentiate the functiony = fl.x) with respect tox to get dy = /(x). dx
dx Second, differentiate the function x = g(t) with respect to t to get-= g'(t).
dt
Third, multiply these two derivatives to obtain the third, dy. An example should dt
clarify this.
6
Example 1 Given: y = x 3 - 3x2 + 4x- 8 = j{x) and x = t 2 + 2t = g(t). Use the chain rule to find:.
Solution dy = 3x2 - 6x + 4 = 3(t2 + 2t)2 - 6(t2 + 2t) + 4 dx
dx dt = 2t + 2 = 2(t + 1)
: = (:)(!)= [3(t2 + 2t)2 - 6(t2 + 2t)+4) 2(t+ 1). Ans.
We could have arrived at this same answer by another route. That is, we could have first substituted the value of x in terms oft into the equation for y, which would give
When we differentiate with respect tot, we get
dy = 3(t2 + 2t)2(2t + 2)- 3(2)(t2 + 2t) (2t + 2) + 4(2t + 2) dt
Factor out (2t + 2) and obtain
dy = [3(t2 + 2t)2 - 6(t2 + 2t) + 4] (2t + 2) dt
which is equivalent to the first answer we obtained.
Since, from the chain rule, we got dr = (:)(:). it is possible to get ( :) by
dividing by (dx), as long as dx =I= 0. Doing so, we get dt dt
dy (dy/dt) -=--dx (dx/dt)
This form is particularly useful when we deal with parametric equations, such as x = g(t) and y = h(t), because it enables us to find the derivative of y with respect to x directly from these equations, as is illustrated in the following two examples.
Example 2
In Art. 5 were given the parametric equations for a trajectory: x = Voxt andy = Voyt- ~ct2.
dy Let us find dx"
Solution dx dy dt = Vox and dt = V Oy - ct
dy (dy/dt) Voy- ct dx = (dx/dt) = ---v;;;-
Substituting fort=~ ax' we get
dy Voy x , - =- -c --=f (x). Ans. dx Vox v5x
This is the same answer we obtained by a different method in Art. 5.
7
Example 3 dy
Given: the parametric equationsx = 2t + 3,y = t 2 - t. Let us find dx"
Solution
then
x-3 or, in terms of x, since t = - 2 -,
dx dt = 2 and
dy -=2t-1 dt
dy dy /dt (2t - 1) dx = dx/dt = -2-
dy X- 4 dx=-2-. Ans.
7. General Expression of Chain Rule
The chain rule is frequently expressed by using letters other than the letter t tore
present the parameter. The most common of these is the letter u. Generally, y is stated
as a function of u and u as a function of x, which, in symbolic form, is written as
Y =flu) and u = g(x)
From the rule we get
~= (du)(:) We have already had an illustration of this formula when we developed the derivative
with respect to x of
y =un
Taking a derivative of both sides of this equation with respect to u, we get
dy -=nun-1 du
Invoking the chain rule, we get
du dx
Self-Test 3
For each set of parametric equations, fmd :.
1. X = 3t2 + 2, y = t2
3. y = u6 , u = 2 + x 2
4. y = u2 + 3u + 8, u = 2x + 3
2 5 y = 2 V2 + - v = (3x + 2)213 . v2'
8
The Derivatives of Trigonometric Funtions
8. Trigonometric Formulas
The following formulas of trigonometry, which should be sufficient for the purposes
of this text, are collected here for convenient reference:
sin A tanA =-
cosA
I secA=-
cosA
I cotA =-
tanA
I cscA =-
sinA
sin2 A +cos2 A= I
sec2 A=I+tan2 A csc2 A = I + cot2 A
sin(A +B)= sin A cosB +cos A sin B
sin(A - B) = sin A cos B - cos A sin B
cos(A +B) = cos A cos B - sin A sin B
cos(A- B) = cos A cos B + sin A sin B
( ) tan A+ tanB tan A +B = ----
I- tan A tanB
sin 2A = 2 sin A cos A
cos 2A = cos2 A- sin2 A
cos 2A = 2 cos2 A - I
cos 2A = I - 2 sin2 A
. A ~-cosA sm-= 2 2
cos:i=. ~ 2 v~
sin A + sin B = 2 sin (A 2+ B) cos (A; B)
sin A- sinE= 2 sin (A; B )cos (A 2+ 8 )
f.A +B) (A- B) cosA + cosB= 2 cos\-2 - cos - 2 -
(A+B) (B-A) cosA-cosB=2sin - 2- sin - 2-
9
9. Circular Measure or Radians
The circle C in' Fig. 2 has a radius of r units. The arc lengths, between lines OX and
OB, is equal to the product r8, where 8 is the central angle, in radians. Thus if s = r8, then
the angle 8 =!..,which defines the angle in circular measure. r
8
c s X
Fig. 2 For the circle C, Arc-length s = r9
In differential form, the equation s = r8 is written as ds = rd8, where ds is the elemental arc length, a() is the elemental angle, and r is a constant.
For emphasis, let us restate the definition of an angle in circular measure:
The circular measure of an angle is the quotient of the length of an arc of a circle
divided by its radius. The radii that subtend the arc are the sides of the angle. Note
that this angle is the central angle () shown in Fig. 2.
Again for emphasis, let us say that the unit of the angle in this measure is the radian.
One radian is the angle for which the arc length s equals the radius of the circle. Any
angle may be said to contain a certain number of radians. Note, however, that the quo
s tient- is dimensionless. It is customary t.:> speak of the angle 8 without using the word
r 2 1T
"radian." Thus, we may speak of the angle 2, the angle 3' the angle 4' the angle 2x, and
so on. During our early school days we learned that the circumference of a circle is 2rr r,
where r is the radius of the circle. At that time little attention was given to the factor 2rr.
Now we must note that 2rr is the number of radians in 360°, rr is the number of radians in
1T 180°, 2 is the number of radians in 90°, and so on. Of course, we should all know that
rr = 3.1415 9 to five decimal places. Usually we are permitted to round it off to 3.1416,
or even to 3.14, where precision is not demanded. In all work involving calculus, all angles which occur are understood to be expressed
in radians. In fact, many of the calculus formulas would be false unless the angles in
volved were so expressed. The student should carefully note this fact, even though the
reason for it is not yet apparent. With this new information about angles in radians, such a trigonometric equation as
y = sinx
10
should take on added meaning. If the unit of x is not specified in degrees, it must be
taken to be in radians. Also, y = sin x may be considered as defining a functional
relation between two quantities exactly as does a simpler equation, such as y = x 2 . For
we may assign any arbitrary value to x and determine the corresponding value of y. This
may be done either by direct computation or by means of a table of trigonometric func
tions, in which case we must interpret the value of x as denoting so many radians, which
again may be converted to so many degrees. Since many tables have angles in degrees, it
may be necessary to convert the value of x from radians to degrees. This can be done 180°
readily by multiplying it by the factor--. Of course, knowing the trigonometric func-. n
. o o o o o nnn tlons of 0 , 30 , 45 , 60 , and 90 , and that these angles in radians are 0, -, -, -, and
6 4 3
n l' is invaluable.
sin (.6.x) 10. Limiting Value of _...o.....-....:.
.6.x
One of the reasons for expressing an angle in circular measure is that it makes it pos
;ible to show that
sin .6.x Lim--= I
~x + o .6.x
which is a very useful relationship, as we shall see. There are a number of ways to prove
that this equation is true. We shall do it by actually calculating the value of sin .6.x. That .6.x
is, we will substitute values for .6.x and sin .6.x and calculate the quotient. The values and
computed quotients are given in Table 1.
Angle .6.x deg
20 10 5 4 3 2 1 %
TABLE 1
VALUES AND COMPUTED QUOTIENTS FOR SIN .6.x AND .6.x
Angle .6.x sin .6.x Radians
0.34907 0.34202 0.17453 0.17365 0.08727 0.08716 o .. o6981 0.06976 0.05236 0.05234 0.03491 0.03490 0.01745 0.01745 0.00873 0.00873
sin .6.x -:;sx-0.97980 0.99496 0.99874 0.99928 0.99962 0.99971 0.99982 1.00000
sin .6.x Observe that .6.x did not have to become very small before -- became very nearly
.6.x
unity. This demonstration clearly indicates that the desired limit is unity.
The table also shows that sin ( x) = x, in radians, approximately, when the value of x is small.
11
11 L. . . V 1 f [ 1 - cos Llx] . amating a ue o Llx
The limit which was established in the preceding article may be used to find
the limit of P -~; Llx J as Llx approaches zero. Squaring both sides of the trigono
metric identity
we get
. A Y 1- cosA Sln - =
2 2
2 sin2 A ·= 1 - cos A 2
Letting angle A = Llx, we get
I - cos Llx = 2 sin2 Llx 2
Then, dividing both sides by Llx,
1- cos Llx
Llx
Llx 2 sin2 -
2
Llx
. Llx . Llx sm2
=---=sm---Llx 2 Llx
2 2
. Llx sm-
Now, we already know from Art. 10 that Lim 2
--=1 Llx
And we also know that
Therefore,
12. Derivative of sin u
t;.x --+o
2 2
Llx Lim sin-=0
2 t:;.;c --+o 2
1- cos Llx Lim --Ll-x--= 0 x 1, or 0
t:.x-+ 0
From Part 1 of this series, remember that if y = j{x), the derivative of the func-
tionj{x) can be defined as the limit
dy Lim Lly Lim f(x + Llx) - j{x) -= -= dx Llx - 0 Llx Llx - 0 Llx
12
Here,j{x) =sin x, so j{x + 4x) =sin (x + 4x). From the list of trigonometric functions,
we see that
sin (A + B) = sin A cos B + cos A sin B
If A= x and B= ~.this identity becomes
So
So
j{x + 4x) = sin (x + 4x) = sin x cos 4x + cos x sin 4x
j(x + 4x) - j(x) = sin x cos 4x + cos x sin 4x - sin x = sin x (cos 4x - 1) + cos x sin 4x
dy _ Lim j(x + 4x) - f(x) _ Lim sin x(cos Ax - 1) cos x sin 4x -- - +-----dx 4x-O 4x Ax-0 4x Ax
Lim cos Ax - 1 Lim sin 4x : Sin X + A _ COS X A
4x-O 4x L.U.-0 ~x
: 0 X Sin X + 1 X COS X
:COS X
From Art. 10, we know that Lim sin Ax = 1 and from Art. 11 we know that Ax-0 Ax
Lim cos 4x - 1 = Ax-0 Ax
Lim -( 1 - cos Ax) = -0 = 0. Using the chain rule, if y = sin u Ax-0 Ax
and u is a function of x, dy = cos u x du. This new formula for calculating derivatives dx dx
can be stated as a rule:
. dy du Rule: If y = sm u, - = cos u x -
dx dx (19)
Formula 19 is the fundamental formula for finding the derivatives ofthe other trig
onometric functions. We shall demonstrate this by developing the formula for the de
rivatives of two other trigonometric functions. Before doing so, let's show by an
example how formula 19 is used.
Example
d" Find Jx when y = 4 sin (2x + 6).
Solution du
One can readily see that here u = 2x + 6, and dx = 2.
13
Therefore dy dx = ( 4) (cos u) (2), or 8 cos (2x + 6). Ans.
13. Derivative of cos u with Respect to x
The first stipulation is that u is a function of x which can be differentiated. From
the study of trigonometry, we have learned that
cos A = sin (90° -A)
Letting A = u and 90° = .!!:: , 2
cos u = sin ( ~ - u) Hence, by formula 19, the derivative
But
Therefore
=cos(~-u)!(~-u)
! (~- u) =- : and cos ( ~- u) =sin u
d . du - COS U = - SID Udx dx
(20)
Observe the similarity and difference in these formulas. The derivative of the sine
function, formula 19, has a cosine function as a factor, while the derivative of the
cosine function, formula 20, has a sine function as a factor, but it has a negative sign.
J:::xample 1
Find: when y = 4 cos3 (2x + 6).
Solution Because the cosine is raised to a power, we will use a variation of formula {6) in Art. 2. In the variation
d n 1(dw) dx (w ) = nw"- dx
Let w = cos (2x + 6) and u = 2x + 6. Then
du = 2 dx
dw . du -= -sm udx dx
= - 2 sin (2x + 6)
Because y = 4w3, we use the variation of formula (6). Thus,
dy = 4 x 3w3- '(dw) dx dx
= [12 cos2 (2x + 6)][-2 sin (2x + 6)]
= - 24 cos2 (2x + 6) sin (2x + 6) Ans.
Example 2
Find: when y = x3 cos 2x.
14
Solution You should recognize that the right-hand side of the given equation is a product and that thcrc-
d du dv fore dx (1111) = v dx + u dx .
du dl' Let u = x3; then dx = 3x2. Let v =cos 2x; thus dx =- 2 sin 2x.
Therefore,
: = 3x2 cos 2x- 2x3 sin 2x. Ans.
Example 3 dy .
Find dx wheny = cos3 2x sin 4x.
Solution Here again we meet a product, and again we will use the formula for the derivative of a product.
Letting u = cos3 2x,
du 2 . ( dx) dx = 3 cos 2x(- sm 2x) 2 dx
=- 6 cos2 2x sin 2x
Also, letting v =sin 4x,
dv ( dx) dx = cos 4x 4 dx
= 4 cos 4x
Substituting in the formula for the derivative of a product, we get
:=(sin 4x) (-6 cos2 2x sin 2x) + cos3 2x (4 cos 4x)
or
: = -6 cos2 2x sin 2x sin 4x + 4 cos3 2x cos 4x. Ans.
14. Derivative of tan u with Respect to x Again the first stipulation must be that u is a function of x that can be differenti
ated. From the study of trigonometry, we learned that the tangent may be expressed as the quotient of the sine and the cosine of the same angle. Expressed in the form of an equation, this is
sin u tanu=-cos u
Since the right-hand side of this equation is a quotient, we must use the formula for the
derivative of !:L. However, to avoid confusion, we will use ~ instead of _vu • v v
Letting z = sin u,
and letting v = cos u,
dz du - =cosu-dx dx
dv =-sin u du dx dx
15
Thus we have
Substituting for z and v, we get
!!_(sinu\= cosu (cosu~)- sinu(-sinu~) dx cos~} (cos u)2
(cos2 u + sin2 u) du - dx - cos2 u
But from cos2 u + sin2 u = 1 and sec u =co~ u' we get
Example 1 dy Find dx, when y = 3 tan~-
d du -tan u = sec2 udx dx
Solution x du 1 Here u = 3; thus dx= 3· Now using formula 21, we get
Example 2
dy 3 5 (X) Find- wheny =-tan - · dx' 5 3
Solution Note here that the tangent is raised to the fifth power. Thus
: =i (5) tan4 (~):tan(~)= 3 [tan4 (~)] [sec2 (~~+ Therefore
15. Formulas for Differentiation of Trigonometric Functions
(21)
The formulas for the differentiation of trigonometric functions, where u represents any function of x that can be differentiated, are grouped here. Note that formulas 19 to 21 have already been dealt with in detail in Arts. 12 to 14.
d . du dxsmu=cosudx
d . du -cos u=-smudx dx
.!£..tan u = sec2 u du dx dx
16
(19)
(20)
(21)
d du - cot u =- csc2 u -dx dx
Jxsecu =secu tanu~
:fx esc u =-esc u cot u ~
(22)
(23)
(24)
The proofs of the last three formulas, 22 to 24, are similar to the proofs already
given. They are shown here to allow you to practice calculating derivatives and
working with trigonometric identities.
Let y = cot u = - 1- = (tan u) -I. tan u
Let y = sec u = - 1- = (cos u)-1• cos u
dy du Then-=- (tan u)-2 x sec2 u-
dy du Then-=- (cos u)-2 x (-sin u)-
dx dx dx dx
cos2 u du =---x--x-
sin2 u cos2 u dx
du =---x-
sin2 u dx
du =- csc2 u
dx
I . I Let y = esc u = -:-- = (sm u)- .
SID U
Then dy = - (sin u)-2 cos u du dx dx
cos u I du =---x--
sin u sin u dx
du = - cot u esc u -
dx
sin u du =--x---
COS U COS U dx
du =tan u sec u
dx
Study each of the following examples and solve the problems in the self-test
to be sure that you can work with the derivatives of the trigonometric functions.
17
Example 1
Given y = esc 3x, find : .
Solution By formula 24,
dy dx = -3 esc 3x cot 3x. Ans.
Example 2
Giveny = 4 sec2 4x, find dx· Solution
By formula 23, dy d dx = 4 (2) sec 4x dx sec 4x
= 8 sec 4x (sec 4x tan 4x) (4)
= 32 sec 2 4x tan 4x. Ans.
Example 3
Given r =a cot 38, find:~.
Solution By formula 22,
dr = a(-csc2 38) (3) = -3a csc2 38. Ans. d8
Example4
Given x = a(8- sin 8), find dx. d8
Solution dx d8 = a(l - cos 8). Ans.
Example 5
Giveny =cos (4x2 + 3x- 8), find:.
Solution dy . 2 d 2 dx = - sm ( 4x + 3x - 8) dx ( 4x + 3x- 8)
=- (8x + 3) sin (4x 2 + 3x- 8). Ans.
[Note that it is customary to place the algebraic or numerical factor in front of the trigonometric
factor.]
Example 6 Given the cycloid whose parametric equations are x = a(8- sin 8) andy= a(l- cr>~ 8). Find the
slope of the tangent to this curve at three points: a) where 8 = Yz1T; b) where 8 = 1r; and c) where 8 = 0.
Solution
We know that the slope of a curve in the X· andy-coordinate system is:~. Since we are not
given y as a direct function of x, we shall find the differentials dy and dx separately first. Then we
shall divide dy by dx to get the slope. Thus, since x = a(8- sin 8),
dx =a (d8- cos 8 d8) = a(l- cos 8) dtJ
18
and also y =a (1- cos 8)
dy = a[O- (-sin 8)] d8 =a sin 8 d8
Therefore
dy a sine d8 dx =a (1- cos 8) d8
But in this equation a and d8 appear in both the numerator and the denominator and thus may be can
celled out to give the simpler expression
dy _ sin8 dx - (1 - cos e)
which we will now use to find the slopes.
a) When 8 = V21T (radians), or 90°, sin 'lz1T = 1 and cos 'lz1T = 0. Hence
dxdy] = 1 ! 0 = 1. Ans. e=Y21r
b) When 8 = 7T(radians), or 180°, sin 7T= 0 and cos 7T= -1. Hence
::]e=1T = 1 _o(-1) =% = o. Ans.
c) When 8 = 0, sin 0 = 0 and cos 0 = 1. Hence
dy] O O h. h . . d . A dx e = 0 = 1 _ (+1) = 0 w 1c 1s m etermmate. ns.
Self-Test 4
Find the derivative with respect to the indicated independent variable for each of the
following:
1. y =cos 6x 2. x = 4 sin 2t
3. y = 3 tan 3x
4. y = sin 2x cos 3x
5. y = sin2 3x cos 2x
6.y = 3x2 - sin Sx 7. y = 112 sec 2x + lf6 tan3 2x
8. r = a(sin 28 + cos 28) 1-sin3 8
9. r=---cos e
10. xy + cotxy = 0
11. Find the slope of the tangent to the curve y = sin x at the point a) where x = ~ 1T and
b) wherex = 2. 12. In a tangent galvanometer, the current! is indicated by the deflection 8 of the needle
as compared with the deflection 8a produced by a known current/a, in accordance with
~e il the expression/=Ia--. Noting thatla and 8a are constants, find-.
tan 8a d8
16. Inverse Trigonometric Functions
The equation x = siny
19
defines a relation between the quantities x and y which may be stated by saying either
that x is the sine of angle y, or that y is the angle whose sine is equal to x. When we wish
to use the latter form of expressing x = sin y, we write instead the equation
y = sin-1 x
Here the - 1 is not to be understood as a negative exponent but as part of a new symbol
sin-1 • To avoid any possible misunderstanding, y = sin-1 xis sometimes written y =arc
sinx. These three equations have exactly the same meaning:
x =sin y y = sin-1 x y =arc sin x
It is necessary to be able to work with either form without difficulty.
In x = sin y, y is considered the independent variable, while in y = sin- 1 x, x is con
sidered the independent variable. This equation then defines a function of x which is
called the inverse sine of x. It will add to the clarity of your thinking if you read ''y =
sin-1 x" as ''y is the angle whose sine is x." Similarly, if x =cosy, th~n y = cos1 x; if x =tan y, theny = tan-1 x; and so on for
the other trigonometric functions. In any of these equations it should be noticed that y is not completely determined
when x is given, since there is an infinite number of angles with the same sine, cosine, or
F 1 . I Th "f 1 1T 51T 137T tangent. or examp e suppose y = sm- x. en, 1 x =- ,y = -,-,-, and so on.
2 6 6 6 This creates a certain amount of ambiguity in using inverse trigonometric functions.
We have met this same situation when we went from the equation x = y2 to the
equation y =±...;X. Here if xis given, there are two values ofy.
In order that sin-1 x may be a single-valued function, it is necessary to limit the
• 1T 1T
range and adopt the convention that only angles between -2 and + 2 are accepted.
Thus y = sin-1 x is the angle, in radians, between-~ and+~, whose sine is x.
Similarly, y = tan-1 x is the angle, in radians, between - i and + ~· whose tangent
isx. The same convention applies to csc-I x, but a slightly different type of defini
tion is required for cos-1 x, sec-1 and cot-1 x. For y = cos-I x,y is the angle, in radians, between 0 and rr, whose cosine is x.
For y = cot-1 x, y is the angle, in radians, between 0 and rr, whose cotangent is x.
For y = sec-1 x,y is the angle, in radians, between 0 and rr, whose secant is x.
17. Derivative of sin-1 u with Respect to x.
If y = sin -I u, where u is a function of x, then sin y = u. Take the derivative of
both sides of the latter equation with respect to x and solve for dx as follows:
sin y = u
dy du cosy-=
dx dx
20
dy du dx =cosy x dX
However, it is necessary to calculate cosy when sin y = u.
Fig. 3 Right Triangle with angle y = sin-! u
Remember that the sine of any angle A is defined in terms of a right triangle
in which one angle measures A. Draw a right triangle with one angle measuring y
as is done in Fig. 3. The sine function is the ratio of the leg opposite the angle to the
hypo~enuse of the triangle. Sin y = u, or~. so let u be the length of the leg opposite 1
angle y. Then the hypotenuse of the triangle will be 1 and, by the Pythagorean
theorem, the leg adjacent to angle y isJ 1 - u2• Then
leg adjacent ~ __ r;----'21 _ u2 cosy= v 1 - u
hypotenuse 1
This value of cos y can be used in the above equation.
dy 1 du 1 du dx = cosydx = ~dx
or d . 1 du
dx sm-•u = ~ dx (25)
Thus we have a general formula for the derivative of an inverse sine function. Example _ d
Giveny = sin-1 v'1=4x2, wherey is an acute angle, in radians, find Jx. Solution -- 2
In the given equation, u = yl - 4x2 and u 2 = 1 - 4x . Apply formula 25 to get
dy I du I - 4x dx- .J I- (1- 4Xl)dx = ,J4?.x ~-
- 2 =~ Ans.
18. Derivative of arc sin (:)
Let y = sin -I(;} Then, by formula 25,
dy I I 1 dx Ax;= .J a2- x2
1-2 a
If you want, you can demonstrate this derivative without using formula 25. The
procedure is as follows: . I X . X
y = sm- -, so sm y =-a a
21
Taking the derivative of both sides of the equation for sin y,
dy I dy I cosy-=-, so-=---
dx a dx a cosy
It is necessary to find cosy. Draw a right triangle with angle y as is done in Fig. 4.
Sin y is the ratio of the leg opposite angle y to the hypotenuse of the triangle. Sin
y =~. so let x be the leg opposite y and let a be the hypotenuse. By the Pythogorean a
theorem, the leg adjacent to y is J a2 - x 2• Then
and
adjacent leg J a2 - x 2
cosy= =----hypotenuse a
dy = ---=-~~=::=- Ans
dx a cos y J a2 - x2 - -J a2 - x2 · ax
a
X Fig. 4 Right triangle with angle y = sin -t-;;
19. Derivative of cof1 u with Respect tox
Let y = cos- 1 u where u is a differentiable function of x. Just as we did in Art. 17,
let us convert the inverse function to cos y = u. The relationship between y and u is
shown by the right triangle in Fig. 5. Note that the side u is adjacent to the angle y here,
y
" Fig. 5 Right Triangle with angle y =cos-t u
though not in Fig. 3. Be careful about this difference. Taking the derivative of both sides
of the equation cosy = u, we get
. dv du - smy ::L. =-
dx dx
. . . dy 1 du DlVlde by- smy to get-=--.- -
dx smy dx
From Fig. 5, siny = v'}7; therefore,
d 1 du -cos-1 u =- -dx ~dx
(26)
22
Thus we have a general formula for the derivative of an inverse cosine function.
Ex=:: .. y = ,., -'(:'). Find : .
Solution 2 In the given equation, let u = :L. App;ying formula 26, we get
a
or
dy 1 d (x2) a 2x dx =- -/0----====2 dx \-;- =-vaL x4 -;;
dy 2x
dx=- vfa2-x4" Ans.
20. Derivative of tan-1 u with Respect to x
Let y = tan-1 u, where u is a differentiable function of x and the relationship be
tween y and u is shown by the right triangle in Fig. 6. Again note that the side adjacent
to the angle y is given the value of unity.
If
then
and
or
u
Fig 6 Right Triangle with angle y = tan -1 u
dy du sec2 y-=
dx dx
tany = u
and
But from Fig. 6, cosy=~ or cos2 y = 1 } u2
Therefore,
dy d ldu - =- tan-1 u = -- -dx dx 1 + uz dx
23
(27)
Example 1
. -l(x + 2) dy Gtven y = tan - 2- , f"md dx ·
Solution x+2 du 1
In the given equation, let u = - 2- ; hence dx = 2 .
Applying formula 27, we get
dy 1 (1) 4 (1) dx = 1 + (X; 2 r 2 = 4 + x 2 + 4x + 4 2
or dy 2
dx = x 2 + 4x + 8 · Ans.
Example 2
1 -1 ( b ) . dy Giveny =-tan -tanx fmd-ab a ' dx"
Solution
In the given equation, let u = (;tan x). Using formula 27,
dy 1 1 d (b ) dx = ab (b ) 2 dx ;tanx
I+ -tanx a
21. Formulas for Differentiation of Inverse Trigonometric Functions
The formulas for the differentiation of inverse trigonometric functions are listed
here. In each formula, u represents any function of x that can be differentiated. Note
that formulas 25 to 27 have been dealt with in detail in Arts. 17 to 20.
d I du - sin-1 u = -dx ~dx
d _1 I du -cos u=- -dx y1-U2 dx
d _1 I du -tan u=--dx . I+ u2 dx
d _ I du -cot 1 u=-~dx I+ u2 dx
24
(25)
(26)
(27)
(28)
Example 1
d -l 1 du - sec u = _ r;:;--; -dx uvu2- 1 dx
d _1 1 du -esc u=- -dx uyur.:-f dx
Giveny = cof1 yx2 + 2x, fmd :.
Solution
and
u =v'x2 + 2x
du 1 2x + 2 (x + 1) dx = 2 (x 2 + 2x)Y2 = yx2 + 2x
Applying formula 28, we get
dy 1 (x + 1) dx =- 1 + (x2 + 2x) y'x2 + 2x
dy (x + 1) 1 dx =- (x + 1)2 (x 2 + 2x)Y. =- -(x-+-1)-(x-,2=-+-2x-)=y..
Example 2
Giveny = csc-1( x; 2 ). find:.
Solution x-2 du 1
u =-2 - and dx =2
Applying formula 30,
Ans.
dy = 1 1 ---=2=-===-
dx ( x; 2) .(<x ~ 2)2 _ 1 2=- (x- 2)yx2- 4x
Example3
(29)
(30)
Ans.
_ 3x - 1 (3) dy dy dx G!Veny = (x 2 + 9) +cot -;,a) find dt; b) fmd the value of dt when x = 3 and dt = 9.
Solution The derivative of the quotient is
.!!_(~)- (x 2 + 9)(3)- 3x(2x)dx _ 27- 3x2 dx dt x 2 + 9 - (x 2 + 9)2 dt- (x 2 + 9)2 dt
The derivative of the inverse cotangtnt function is
! oof' (~ )·- 1 + ~~)'(~~): •- (<;'+ 9) (- ;.):
3 dx = x 2 + 9 dt
25
Combining the two terms, we get
dx Whenx = 3 and dt = 9,
dy [ 27 - 3x2 3 J dx dt= (x2 +9)2 +x2 +9 dt
27- 3x2 + 3(x2 + 9) dx (x2 + 9)2 dt
54 dx = (x2 + 9)2 dt. Ans. (a)
d~l 54 3 dt Jx=3 = (9 + 9)2 (9) = 2· Ans. (b)
Self-Test 5
Find: in each of the following:
1. y = sin-1 2x 2. y=tan- 1 (2x- 1) 3. y = cos-1 3x
d8 Find- in each of the following:
dx
7. 8 = sin- 1 ( 1 ) 8. 8 = ..!_ tan-1 ..!_ x
3 3
4. y = sec 1 3x 5. y = csc-1 (2x + 1) 6. y = coC 1 (x2 - 1)
9. 8 = tan-1 (x- 1) x+1
10. 8 = x sin- 1 x + .y'f+X2
The Derivative of Exponential and logarithmic Functions
22. What Is an Exponential Function?
An exponential function is one in which a variable appears in an exponent. Thus d",
52x, x2x, ex 2 , and alog x are exponential functions. Quantities such as x 5 and (x 3 + 2x)2
are NOT exponential functions, because the exponents are constants. They are
algebraic functions. We may do well to refresh our understanding of the properties of exponential func
tions. Let's look at the equation y= bX
where b is any positive constant. If x = n, an integer, then y is determined by raising
b to the nth power by multiplication. If x =!!..., a positive fraction, then y is the qth q
root of the pth power of b, or the pth power of the qth root of b. In symbols, p - q
y = bq = ::;t;P And if X=- m y = b-m = - 1- .
' bm Furthermore, if x = 0, y = b0 = 1.
To illustrate these properties, let us take an equation such as y = bX and plot it on a
graph. We will assume that b = 1.5; then we will have the equation y = (1.5)X. A number
of positive and negative values have been assigned to x. The values were calculated and
tabulated as shown in Table 2, and plotted in Fig. 7. Observe that the curve in Fig. 7 is continuous for negative values of x as well as for
positive values of x, and that y is positive for all values of x. When x is a large positive
26
y
8 I 6 I 4 I
/ 2 ./
·" ·"'
·-· -·-· X
-4 -2 0 2 4 6
Fig. 7 Graph ofy = (l.S)x
TABLE 2
DATA FOR PLOTTINGy = (1.5)x
X y = (1.5)X
5 7.59375 4 5.0625 3 3.3750 2 2.2500 1 1.5000 % 1.2247 '4 1.1067 0 1.0000
-v.. 0.9036 -% 0.8166 -1 0.6667 -2 0.4444 -3 0.2963 -4 0.1975
value, y is a larger positive value. When x takes on large negative values, the curve ap
proaches and is asymptotic to the line y = 0 or to the x-axis. When x is zero, y is
unity. When b is greater than unity, the graphs of all equations of the form y = bx
have a similar shape.
27
23. Logarithms
If a number x can be obtained by computing the result of placing an exponent
y on another number a, then y is said to be the logarithm of x to the base a. That is, if
X= aY
then y = loga x
These equations are simply two different ways of expressing the same fact concerning the
relation of x, y, and a. You should make yourself so familiar with these forms that you
can use one or the other as convenience may demand. From these equations, you can easily prove the fundamental properties of loga
rithms. These are stated below in equation form as a reminder.
n IogaP = Ioga pn
Ioga I = 0
I log -=-log P a p a
Since you have had considerable experience with logarithms to the base I 0 [a= 1 0],
you should use this base to prove the preceding equations, and at the same time refresh
your knowledge of logarithms. When you are doing this, we suggest that you use easy
round numbers, such asP= I 000 and Q = I 00.
24. Base of Natural Logarithms
Theoretically, any positive number, except 1, may be used as the base of a sys
tem of logarithms. Practically, however, there are only two numbers so used. The
first is the number 10, which is the base of the common system of logarithms. Com
mon logarithms are very convenient for calculations and are used almost exclusively
with trigonometry. Another number, however, is more convenient in theoretical developments, since it
gives simpler formulas. It occurs often in problems in engineering and physics. This num
ber is denoted by the letter e. Frequently, it is written as e, the small Greek epsilon. Un
less a statement is made to the contrary, you are to understand that e (or e) is used to
denote the base of natural logarithms. These are sometimes referred to as Napierian or
hyperbolic logarithms. The value of e may be expressed by the infinite series
1 I I 1 1 e=I+-+-+-+-+-+
I 2! 3! 4! 5!
where 2! = 2 X 1, 3! = 3" X 2 X I, 4! = 4 X 3 X 2 X I, 5! = 5 X 4 X 3 X 2 X I, and so on.
Computing the infinite series to seven decimal places, we have
e = 2.71828I8 ...
28
For most practical purposes, however, it is sufficiently accurate to take
e = 2.7183
Since logarithms to the base e are called natural logarithms, we shall denote them by
the symbol Ln, wP,ere the letter L denotes a logarithm and the letter n denotes natural.
Thus the symbol Ln should be easy to remember. However, other texts may use the
symbol loge instead of Ln. The symbol log without any base indication will be taken to
mean a logarithm to the base 10. From the definition of a logarithm, we should know that if
X: eY
then y: Ln X
Caution and attention are very necessary when using tables and other texts. The
symbol for the natural logarithms may be written as log, and it will rest upon the reader
to know that the natural logarithm is meant. A good general rule to adopt is: Whenever
a logarithm appears in calculus, it is a natura/logarithm. A useful relationship that exists between logarithms of different bases is
Lnx log 10x =-- = log 10 e Ln x
Ln 10
from which we can gather that if .x = e, then
1 log 10 e=-
Ln 10
The above relationships can be expressed more generally as
Lnb log a b = -- = log a e Ln b
Lna
Many calculators have a key for the natural logarithm and one for the common
logarithm of any positive number. Tables of natural logarithms exist and should
be used if available. For example, we can find from such a table that Ln 10 = 2.30259, or in a table of common logarithms log e = log 2.7183 = 0.43429, a:ad by simple
1 arithmetic, 2_30259 = 0.43429.
25. Derivative of ex
There are a number of ways to find the derivative of ex. Perhaps the simplest way is
to write ex as a power series and then to differentiate every term in the series.
In Art. 24, e was expressed as a series. Here it is obvious that e has the exponent I.
When the exponent of e isx, then the series is
x2 x3 x4 xs ex= 1 +x +- +- +- +- +
2! 3! 4! 5!
Taking the derivative with respect to x, we get
d 2x 3x2 4x3 5x4 -(ex)=O+ I+-+-+-+-+ ... dx 2! 3! 4! 5!
29
but
Hence
2 -= 1• 2! '
d x 2 x 3 x4 x 5 -(ex)= l+x+-+-+-+-+ ... dx 2! 3! 4! 5!
d Note that the right-hand sides of dx (ex) and ex are identical. Thus the derivative of
ex in series form is a series which is itself the series form of ex. That is,
which is unique! This is the only case in which the derivative of a function is equal to the
function itself! d
Note that since the derivative- (ex)= ex, the curve y =ex cannot have any turning dx
points-that is, no maxima or minima-because ex cannot be zero. The curve y =ex will
have a shape similar to the curve shown in Fig. 7.
26. Derivative of ebx
Let y = e bx, where b is a constant. Also, let z = ex; then z b = e bx, or y = z b. Now
we have y as a function of z and z as a function of x. Applying the chain rule gives us
dy dy dz -=--dx dz dx
Here dy - = bzb-1 dz
Also, since z = ex,
Hence
But
Thus dy -- = be bx e x ex = be bx dx
Therefore,
(31)
27. Derivative of eu
Let y = eu, where u is a differentiable function of x. Again using the chain rule, we
get dy = dy du
dx du dx
30
where
Therefore, d du -(e") = e"d--: dx
This is a very useful formula.
Example 1
. - 3x2 dy G1veny = e , fmd ax·
Solution
Apply formula 32.
dy -3 2 d -3 2 -=e x -(-3x2)=-6xe x Ans. dx dx
Example 2
Giveny = e-3x sin 2x, find:.
Solution
dy -3x d -3x d - = e - (- 3x) sin 2x + e cos 2x-(2x) dx dx dx
=- 3e- 3x sin 2x + 2e- 3x cos 2x
= e-3x (2 cos 2x- 3 sin 2x). Ans.
Example 3
Giveny = sin- 1 (::: :=:). find :.
Solution d _1 1 du
Here we use dx sin u = y' 2 -, in which 1- u dx
and
4
31
(32)
Therefore,
Self-Test 6
Find : in each of the following:
1. y = e2fx 2. y = e-x2
5. y =x2 e3x
6. ex+ eY = 1 3. y = (2x2- 2x + 1)e2x 7. ex siny = 2
4. y=(3cosx+sinx)e 3x 8 sin x + cosx . y = ex
28. Derivative of Ln bx.
By the definition of a logarithm, if y = Ln bx, then bx = eY. Take the derivative
with respect toy and get
or
Hence
But y = Ln bx; therefore,
dx d b- =-(eY) =eY = bx
dy dy
dx -=x dy
dy =
dx X
dy d 1 - = - (Ln bx) = - = x- 1
dx dx X (33)
This result is very important in engineering and physics. It is also interesting to note that
it fills a gap that has existed until this moment. You will recall that we have derived and
d . 1 used the formula dx (xn) = n xn- , where n is any number, positive or negative, integral
or fractional. For example, we have found that
d dx (x2) = 2x
d - (x) = x 0 = 1 dx
32
and so on. From the preceding equations, observe that there is no power of x which,
when differentiated, will give an expression equal to x-1. In this article we have finally
found that this missing link is Ln x, whose derivative is 1/x or x-1•
29. Dtmvative of Ln u
Let y = Ln u, where u is a differentiable function of x. From the definition of a
logarithm we know that
u=eY
By the application of formula 32, the derivative with respect to x is
du d dy - =-(eY) =eY-dx dx dx
dy dy 1 du Solving for- we get-=--. Buty = Ln u and eY = u; therefore,
dx' dx eY dx
Example 1 3 dy
Given y = Ln 2x , find dx.
Solution
d 1 du - (Ln u) =-dx. u dx
Here u = 2xl and we apply formula 34 to get
Example 2
Find:, givenxy =Ln (x + y).
Solution
(34)
Since this is an implicit function, take the derivative of both sides of the equation with respect
to x; thus
Transposing and factoring, we get
Solving for:.
dy 1 ( dy) x-+y=-- 1+-dx (x + y) dx
( 1 )dy 1 x- (x + y) dx = (x + y)- Y
dy 1- y (X+ y)
dx = x (x + y)- 1· Ans.
E_xampl~ 3 3 . dy . . . . . . . . G1ven y = Ln 2x , fmd dx' but s1mphfy the loganthm1c expressiOn before d1fferentJatmg.
33
Solution Here we write Ln 2x3 as Ln 2 + Ln x 3. Remembering that Ln x3 = 3Ln x,y = Ln 2x3 becomes
y = Ln 2 + 3Ln x and
: = 0 + 3(~ )= ~. Ans.
Because Ln 2 is a constant, its derivation is 0. Compare this result with the answer for example I.
Example4
Find:. given y = Ln (l- x 2)213.
Solution 2
Let u =(I - x2)3. Then
du 2 _.!.. d 2 --'--=-(1 -x2) 3-(1 -x2) =-(I -x2) 3 (-2x) = dx3 dx 3
and
Ans.
30. Derivative of au
Let y =au. It is understood that a is a constant and u is a differentiable function of x. Taking the natural logarithm of both sides, we get
Ln y = Ln au = u Ln a
Differentiating with respect to x, we obtain
or
Therefore,
Example 1 2 dy
Given y = a3x , find dx .
Solution
1 dy du du - - = u 0 + Ln a - = Ln a -y d.x d.x d.x
dy du -=yLnad.x d.x
d du - (a")=a" Ln adx dx
~ ~ 2 Let u = 3x 2, then dx = 6x. Then apply formula 35 to get dx = 6x a3x Ln a. Ans.
Example 2
Find dy ify = x 2 3x dx .
Solution
(35)
We should recognize that y is the product of two factors. The derivative of x 2 with respect to x is 2x, and the derivative of 3x with respect to xis 3x Ln 3. Thus
dy d d -= 3x_(x2)+x2-(3x) dx dx dx
= 2x 3x +x 2 3x Ln 3. Ans.
34
Etamp/1.'3
.. dy tan- 1 x f-md- when v = 10 · ·
dx ·
Solution -I du I
Here we should recognize that a= 10 and u =tan x. Furthermore,- = --2 . according to dx l + x
dl· I 0 tan -I x . formula 27. Therefore, applying formula 35 we get--"- = l.n I 0. Ans.
' - dx I + x 2
31. Derivative of Loga u Let y =log au, where a is any positive number except 1 and u is any differentia
ble function of x. Since we have already developed a rule for the derivative of a natural logarithm, we should try to convert loga u to a natural logarithm. We do this by applying
Lnu the rule mentioned in Art. 24. Hence loga u = L- = loga e Ln u. Here Ln a and loga e na are constants; Ln u is the variable. Therefore, differentiating each of these expressions, we get
Hence,
Example
dy d 1 1 du 1 du -=-(log u)=---- =log e-dx dx a Ln a u dx a u dx
..
~(loga u) =_I_ du = loga e du dx uLna dx u dx
dy Given y = log sin x 2, find dx.
(36)
Solution In this equation the base of the logarithms is 10 and u is sin x 2. Applying equation 36, we
get
By formula 19,
Therefore,
dy loge d 2 - = -- - (sin x ) dx sinx 2 dx
dy loge dx = ----2 (2x) cos x 2 = 2x (log e) cot x 2. Ans.
smx
or, in terms of natural logarithms,
dy 2x cotx 2
dx - Ln 10
32. Using Logarithms to Simplify Differentiation
Ans.
The natural logarithm may be used to considerable advantage in two types of expressions to simplify the work of obtaining the derivatives. The first type is an expression in the form of products, quotients, or a combination of products and quotients. The second type is an expression in which a variable has a variable for an exponent. The following two examples should help to clarify what we mean.
35
Example 1 /ex- l)(x- 2)
Differentiate y = V (x _ 3) (x _ 4) .
Solution Take the natural logarithm of both sides and get
Ln y = 1/2 [Ln (x- 1) + Ln (x- 2)- Ln (x- 3)- Ln (x- 4) J
Take the derivative of both sides with respect to x.
~ ~ = + [x ~ 1 + X ~ 2 -X ~ 3 -X ~ 4] 1 [ -2 (2x 2 - lOx+ 11) ]
= 2 (x- l)(x- 2)(x- 3) (x- 4)
dy (2x 2 -lOx+ ll)y -=-dx (x-l)(x- 2)(x- 3)(x-4)
(2x 2 - lOx+ 11) l(x- 1) (x- 2) (x- l)(x- 2)(x- 3)(x- 4) (x- 3)(x- 4)
2x 2 -10x+11 Ans.
Example 2 x Differentiate y = xe .
Solution Taking the natural logarithm of both sides, we get
X
Ln y = Ln x e = ex Ln x
Now we have a product which is easily differentiated. Doing so, we obtain
-- = ex - +ex Ln x = ex - + Ln x 1 dy 1 (1 ) y dx X X
dy (1 ) -=yex - +Lnx dx x
or
33. Graph of y = Ln x
Since we will be dealing with many expressions involving the logarithmic function, we should analyze this important and simple function to learn some of its characteristics. Let us graph the curve for y = Ln x. To do so we select a number of values for x (remember that x must be positive) in the equation y = Ln x, and find the corresponding values of y. Doing this will give us a number of pairs of values for x andy. These are shown in
Table 3. Study this table carefully. Since the slope of the curve y = Ln xis given by dy = dx
1 -, we have tabulated the slope at the points we selected. When these points are plotted X
and connected by a smooth curve, we have a graph such as shown in Fig. 8.
36
TABLE 3
DATA FOR PLOTTINGy =Ln x
X !4 !6 1 2 4 6 8 10
y -1.386 -0.693 0 +0.693 1.386 1.782 2.079 2.303
dy 1 1 1 1 4 2 1 -
dx 2 4 6 8 10
Observe from Fig. 8 that the slope is positive for the limited portion of the curve
shown. However, if we analyze the equation of the slope of the curve,
dy 1 -=-dx X
we readily conclude that for all positive values of x the slope, : , will be positive. Hence
the graph of y = Ln x must rise steadily from left to right. Also, since the derivative is
continuous, the function Ln x is itself continuous, and the curve has a continuously turn
ing tangent. Let us look. at the second derivative. Its equation is
d2y 1 dx2 =- x 2
y
3
slope= m 2 ·~
Lna Ln 10
X
0 4 6 8 10
-I
-2
Fig. 8. Graph of Curve y = Ln x
37
Note that for any value of x, positive or negative, the second derivative is negative. As we
have learned, this means that the curve's slope is everywhere decreasing. This fact
substantiates our earlier statement that the curve has a continuously turning tangent.
Observe also that the curve passes through the point ( 1, 0) because Ln 1 = 0. At this
point its slope is + 1, and so the tangent line at this point makes an angle of 45° with the
x-axis. Similarly, the slope at x = 4 is lf4 , and the slope at x = 10 is lfto, and so on.
As a final observation, we note that as x increases without limit, so the value of
y = Ln x increases without limit. That is,
Lim (Ln x) = oo x~oo
1 On the other hand, as x approaches zero through positive values, -will approach positive
X
1 infinity, but Ln x, which equals- Ln -,approaches- 00• Therefore, the curve reaches to
X
negative infinity as x approaches zero from the positive end.
34. Formulas for Exponential and Logarithmic Functions
The formulas for the differentiation of exponential and logarithmic functions are ac
cumulated here for ready reference. In each formula u represents any function of x that
can be differentiated.
d _ (ebx) = bebx dx
d ( u) _ u du - e -e -dx dx
d 1 - (Ln bx)=- = x- 1
dx X
d 1 du dx (Ln u) =-;; dx
d du -(au) =au Ln a -dx dx
d dx log.z u
log.z e du du· =---=----
u dx uLna dx
Self-Test 7
Find: for each of the following:
1. y=Ln(x2 + 2x)
2.y=(Lnx)3
4. y =Ln cosx
5. y = Ln (tanx + secx)
2 3. y =log- 6. y=Ln(x+vX2+9')
x
38
(31)
(32)
(33)
(34)
(35)
(36)
x2 7. y=Ln 1+x2
8. y =x2 Lnx2
X 9. e<x + Y) = Ln
y
11. xy = Ln (x + y)
12. y=Ln(Lnx)
13. y = 3sec x
10. y =Ln .J}+X2- X tan- 1 X 14. y =xLn x
Applications of the Derivative
35. Polar Coordinates
When determining the position of a point in a plane, we have been using two dis
tances, x andy. The x-distance is perpendicular to they-axis, while they-distance is per
pendicular to the x-axis. Thus we have the commonly called rectangular coordinates.
There are many other methods for determining the position of a point in a plane. A
very widely used method uses a distance, called the radius vector, and a directional
angle, called the vectorial angle. The radius vector is measured from a fixed point called
the origin, or the pole. The direction of the radius vector is measured by the angle the
radius vector makes with a fixed line, called the initial line, or polar axis. The distance
from the pole and the angle from the polar axis constitutes a system of coordinates re
ferred to as polar coordinates. Study Fig. 9, in which we have the point P at a distance r from the origin 0 at an
angle () measured from the polar axis OX. Here we see that r and 8 completely describe
the position of point P. They are called the polar coordinates of point P.
Fig. 9. Position of Point P Described by Polar Coordinates r and ()
The quantities r and 8 may be either positive or negative. By accepted convention,
the positive angle is measured in a counterclockwise direction, as shown in Fig. 9. Hence
a negative value of 8 is laid off in a clockwise direction. Positive values of r are measured
from 0 along the terminal line of 8. Negative values of r are measured from 0 along the backward extension of the terminal line.
Let us examine Fig. 10 to clarify the meaning of negative and positive polar co
ordinates. Observe that the point P 1 has the coordinates 6 and 60°. Both the radius
39
~ (G ,Go•J
~ (-5,-30°)
X
~(5,-30°)
~ (-G ,GO•)
Fig. 10. Examples of Points with Negative and Positive Polar Coordinates.
vector and the directional angle are positive. The point P2 has the coordinates ( -6, 60°).
Here the radius vector is negative, and is therefore measured from the origin along the ex
tension of line OP 1 until it reaches point P 2•
Point P3 has a negative angle and point P 4 has a negative angle and a negative radius
vector. The point P4 could also be described as a point at(5, 150°) andP2 as a point at
( 6, 240°). Furthermore, P2 could also be described as a point at ( 6, -120°). From these
few examples, it should be obvious that the same point may have more than one pair of
coordinates, which is somewhat different from the rectangular coordinate system. In
practice, it is usually convenient to restrict r and () to positive values.
An equation using polar coordinates may define the locus of a point, and may be
plotted by adhering to the meaning of polar coordinates. Thus the equation r = 4 defines
a circle of radius 4 and the center at the origin, because the locus of all points at a dis
tance 4 from the origin and in a plane is a circle. This is very similar to what we have
already met in rectangular coordinates. For example, the equation y = 4 defines a
straight line four units from the x-axis. In a polar coordinate equation using () but not involving a trigonometric func
tion of fJ, it should be understood that() is always expressed in radians. For example,
to graph r = 8, the length of the radius vector at each point will be the value of fJ in
radians. When () = 180°, r will be rr or 3.14 and when fJ = 360°, r will be 2 rr, or 6.28.
36. Graphs with Polar Coordinates.
When an equation is given in polar coordinates, the corresponding curve may be
plotted by giving the angle () convenient values, computing the corresponding values of
the radius vector r, plotting the resulting points, and drawing a curve through them.
Plotting in polar coordinates is easier if you use paper ruled as in Figs. 11 and 12.
This kind of paper is called polar coordinate paper. The angle fJ is determined from the
numbers at the ends of· the straight radial lines. The value of r is counted off on the con
centric circles, either toward or away from the number which indicates fJ, according as r
is positive or negative.
40
Example 1
Fig. 11. Graph of a Cardioid Determined by Equation r = 1.5 + 1.5 cos 8
Plot the locus of points given by the equation r = 1.5 (1 +cos e).
Solution To plot this curve, we assign values toe, and compute the corresponding values of r with the aid
of a table of natural cosines or of a calculator. Plot those points of the curve whose coordinates are thus
determined. Proceeding in this manner, we see. that as the values assigned to 8 increase from 0° to 90°, cos 8
decreases from 1 to 0; hence r decreases from 3.0 to 1.5 and we draw the corresponding arc ABC as
shown in Fig. 11. As 8 increases from 90° to 180°, cos 8 decreases from 0 to -1, r decreases from 1.5
to 0, and we draw the arc CDO. As 8 increases from 180° to 360°, cos 8 increases from -1 to +1,
causing r to increase from 0 to 3 in a continuous manner. We draw the arc OEFGA. If any more
values are assigned to 8, the corresponding points will follow the curve already drawn. Hence the
curve in Fig. 11 is the complete curve of the given equation. This curve is called a cardioid.
If we had t2kcn negative values of 8, the curve would have been traced in the reverse direction.
Example 2
Plot the locus of points given by the equation r = ( ~) 8.
Solution When plotting this curve, remember to consider 8 in circular measure (radians). When 8 is zero,
r = 0. As 8 increases, r also increases, so that the curve winds an infinite number of times around the
origin while receding from it, as shown in Fig. 12. For negative values of 8, the curve spirals in the re
verse direction, as indicated by the broken curve. This curve is called the spiral of Archimedes.
41
Fig. 12. Graph of a Curve Determined by
Equation r = (~ J 8
Plot the following curves:
1. r = 4 cos 38 8
2. r = 2 sin 3
37. Polar Equation of Straight Line
Self-Test 8
3. r = 2 + 3 cos 8
4. r = 4 sin 28
Before we leave the subject of polar coordinates, we should derive polar equations
for some common geometric figures. First, let us try to derive the equation for a straight
line. Let the straight line LM in Fig. 13 be drawn perpendicular to the polar axis OX
at A, where OA =a. Let P(r,8) be any point on the line LM. Then OP = r is the hy
potenuse of the right triangle OPA. By trigonometry,
or
Written in another form, we get
OP cos 8 = OA
r cos 8 =a
a r=-
cos 8
This is the equation for a straight line perpendicular to the polar axis.
42
L
P(r,())
r
X
0
M
Fig. 13. Locus of All Points on Line LM Is
r = _!!:___e, the Polar Equation for a Straight cos
Line
a Let us test r = -- for its validity as an equation of a straight line. If we let 8 in-
cos e crease from 0° to 90°, cos 8 decreases from 1 to 0, which will cause r to increase from a
to oo, tracing out the lineAL. As 8 increases from 90° to 180°, cos 8 decreases from 0 to
-1. Thus r is negative. As r numerically decreases from -oo to a, it traces out the line MA.
If any other values are assigned to 8, no new points will be found. Therefore, we have
a established that r = -- is the equation of a straight line. One should note that the part
cos e AM of the line may also be found by letting 8 vary from 0° to -90°.
Of course, we should realize that the equation of this line in rectangular coordinates
isx =a.
38. Polar Equation of Circle
In Art. 35 we mentioned that the equation
r= 4
is a polar equation of a circle having a radius of four units with its center at the pole, or
origin. This same circle in rectangular coordinates is defined by the equation
vx 2 + y 2 = 4
Let's take a circle whose center is not at the origin but is on the initial line, b units
from the origin, and which passes through the origin, such as shown in Fig. 14. Since the
radius of the circle is b units, then OA = 2b. Also, let P(r,O) be any point on the circle.
Line segment OP = r. The triangle OPA is inscribed in a semicircle. We know that
triangle 0 P A is a right triangle, and that
OP= OA cos 8 or
r = 2b cos e which is the polar equation of the circle shown in Fig. 14.
43
c
X
A
Fig. 14. Locus of All Points on Circle C, Described by r = 2b cos()
Let us trace the curve, using r = 2b cos (). As the angle () is increased from 0° to
90°, r decreases from 2b to 0, and the upper half of the circle is constructed. As() is in
creased from 90° to 18.0°, r is negative and decreases from 0 to - 2b, and the lower half of
the circle is constructed. Note that a negative r at an angle in the second quadrant ac
tually places the point in the fourth quadrant. If any more values are assigned to(), the
points located will be the same as those already located.
It should be interesting to compare the preceding equation with the corresponding
one in rectangular coordinates, which is x 2 + y 2 = 2bx
Note the simpler form of the polar equation compared to the rectangular equation.
39. Length of an arc In Part 2 of this series, when we needed to find the length of a curve, we de
veloped the relationship ds2 = d.x2 + dy2
where ds is the differential arc length, and dx and t~r are the differentials in terms of
the rectangular coordinates. As you know from analytic geometry, you can transform rectangular coordi
nates to polar coordinates by using the following equations:
x = r cos 0 and y = r sin ()
With polar coordinates, both r and 8 are variables, so you can differentiate
d.x = cos 8 dr- r sin 0 d8
and dy = sin 8 dr + r cos 8 d8
Squaring both sides of these equations and substituting into ds2 = d.x 2 + dy 2, we obtain
ds2 = dx2 + dy2 =(cos 8 dr- r sin 8 d8)2 +(sin 8 dr + r cos 8 d8)2
= ( cos2 8 + sin2 8) dr2 + ( -2r sin 8 cos 8 dr a8 + 2r sin 8 cos 8 dr d8) +(sin28 + cos20)
r 2 d 82
ds2 = dr2 + r2 d 82 (37)
Study the following example to see how the formula can be used.
Example Find the length of the curve described by the equation r = fi that falls in the first quadrant.
Solution In the first quadrant, () will be between 0 and}-. First calculate dr:
dr = 2() d()
44
· Then calculate ds
ds = .J d? + r2 dff
ds = .J (26 d6)2 + 64 d62
ds = e,J4+82 d6
To calculates, find the integral of both sides of the equation
Let u = 4 + ff. Then du = 26 d6
I .!. s ="2Ju2du
I ~ =- u2 + C
3
I 3 =-(4 + ff-)2 + c
3 8
When 6 = 0, s = 0, so C = -3
3
s = f 6 ...;-;::;:er d 6
I - 8 When 6 :!!:, s = -2 ( 16 + 1r2) 2--. or about 2.82 units Ans.
2 4 3
40. Angles of Tangent Line There is an important relationship that can be observed from Fig. 15. Because
an exterior angle of a triangle is equal to the sum of the nonadjacent interior angles, from triangle OA P,
ct>= e + 1/1 (38)
where ¢ is the angle between the polar axis and the tangent line, () is the angle between the polar axis and the radius vector, and 1/J is the angle between the radius vector and the tangent line.
In rectangular coordinates, when we want to discuss the direction of a curve at a point, we use the angle ¢ from the positive x-axis to the tangent line. In polar coordinates, it is more convenient to use the angle 1/J, which is the angle from the extension of the radius vector to the tangent line T.
0
I I I I I I I I
8
r = f{9)
- P(r, 0)
Fig. 15. Relationship of Angles 0, 0 + 0
45
X
dy From the rectangular coordinate system we know that tan 1> = dx· From the
triangle OBP in Fig. 15, we obtain tan(} = 2:. because x = OB andy= BP. To calcux
late a formula for tan 1/1, use the equation cf> = 0 + 1/J, or 1/1 = cf>- 0. Then use the
formula from trigonometry
tan (A _B)= tan A- tan B 1 +tan A tan B
tan cP - tan 0 tan 1/J = tan ( 1> - 0) = ..,.------
1 + tan cP tan 0
S. dy mce tan 1> = dx
sin 0 dr + r cos 0 d 0 sin 0 ------.---and tan 0 = --. cos 0 dr -r sm 0 dO cos 0
Then, sin 0 dr + r cos 0 d 0 sin 0
cos 0 dr - r sin 0 dO cos 0 tan 1/1 = . 0 . 0 d dO sm (sm r + r cos 0
1 +cos 0 (cos 0 dr - r sin 0 dO)
sin 0 cos 0 dr + r cos2 0 d8 - sin 0 cos 0 dr + r sin2 () dO
tan 1/1 = cos 2 0 dr - r sin () cos () d() + sin2(J dr + r sin 8 cos() d8
r (cos2 () + sin2 0) dO tan 1/1 = .
( cos2 0 + sm2 O) dr
r d() tan 1/1 =--
dr (39)
To calculate a workable formula for tan!f>, use the fact that!f> = () + 1/J and the formula
tan A+ tanB tan (A + B) = as follows:
1- tanA tanB
tan (J +tan -,JJ tan <1> = tan ( (J + -,JJ) = -------'--
1 - tan (J tan -,JJ
The following examples will illustrate how these formulas can be used.
Example 1 a) Find the angle -,JJ in terms of the angle (J for a cardioid having the equation r =a (1 - cos £J),
as shown in Fig. 16. Also. find the angle that the tangent line T makes with the horizontal line OX,
b) when (J equals 120° and c) when (J equals 150°.
Fig. 16. Cardiod Having EqU4tion r = a(l -cos £J)
46
X
Solution a) From the equation of the curve we can find dr =a sin 8 d8, which we can substitute in !lqua
tion 39 and get ,,,_rd8 _a(1-cos8)d8
tan 'I'- dr - a sin 8 d8
(1 - cos 8) Ans. sin 8
b) When () = 120°, cos() =cos 120° =-cos 60° = -T· and sin ()=sin 120° = sin 60° = v;
So 1/1 = tan-1 .J3 = 60°
Then,
(+.!. I -cos() 2
tan 1/1 = = ,-:;3 sin() y _, -2-
So, when() = 120°, the tangent line is parallel to the polar axis.
c) When()= 150°, cos()=- .J3 and sin fJ =J... 2 2
tan 1/1 = I - cos () sin fJ
lli
I
2
2
So 1/1 = tan- 1 3.73205 = 75°
Then.
= 2 + v'3 = 3.73205
Ans.
<P = () + 1/1 = 150° + 75° = 225°
The angle between the polar axis and the tangent line could be expressed as 225°. However, it should
be expressed as the smallest angle between the polar axis and the tangent line. Thus,
<P = 225° - 180° = 45° Ans.
<P = 45° because the tangent line is not a directed line. To say that two lines intersect at an angle of 45° is
the same as saying that they intersect at an angle of 225°.
Example 2 Given the equation of a curve r = 3 sin 38, which is illustrated in Fig. 17, fmd a) the angle as a
function of 8 that the curve makes with its radius vector; b) the slope of the tangent when e = 135°;
c) the angles 8, where the tangent is horizontal.
Solution a) From the equation of the curve we find
dr = 9 cos 38 d 8
The angle between the radius vector and the curve is 1/J and tan 1/J = r ::. Substituting for dr, we
obtain 3 sin 38 d8
tan 1/J = 9 cos 38 d 8
Dividing out the common factors, we get
or
sin 38 1 tan 1/J=--- =-tan 38
3 cos 38 3
l/l=tan-1 (113tan38]. Ans.
b) The slope of the tangent line is equal to tan</>, and
tan8+tani/J tan</>=1-tan8tanl/l
When 8 = 135°, tan 8 = -1; then 38 = 405°, tan 405° = tan 45° = + 1. Thus tan 1/J = 113 tan 38 = 113.
47
90°
120° Go•
240• 3oo·
270°
Fig. 17 .. Rose of Three Leaves Described by Equation r = 3 sin 38
.-: · Substituting these values into the equation for tan 1/J we get
c) When the tangent line is horizontal, tan </J = 0; thus
tan f! +tan 1J; · =0
1- tan e tan "'
Ans.
Substituting If3 tan 3e for tan lj;, we obtain
tan e + lf3 tan 3 f} ----"'---- = 0 1 - lf3 tan e tan 3 f}
Now our job is to find the angle f! that will satisfy this equation. It should be obvious that this equation will be satisfied when we find a value of f! that will make the numerator equal to zero. Setting the numerator equal to zero, we obtain
tan f! + If3 tan 3 f! = 0
To solve for f! in this equation it will be necessary to convert tan 311 into terms of tan e. We can do this as follows:
But
tan 2e +tan e tan (3e) = tan (2e +e) = 1 - tan 2e tan e
2 tan e tan (2e) =tan (e +e)= 1- tan2 e
48
Thus we get
2 tan e 1- tan2 6 +tan 6
tan 3e= 2tane 1-1-tan2etane
3 tan e- tan3 e 1- 3 tan2 e
Substituting this into tan 6 + 113 tan 36 = 0, we have
1 [3 tan e- tan 3 6] 0 tane+- = 3 1-3tan2e
Multiplying by the denominator and collecting terms, we obtain
2 tan e- 10f3 tan3e = 0
Factoring out 2 tan 6, we get
Now we have a product of two factors equal to zero and the rest is simple algebra. Thus 2 tan 6 = 0;
also, 1- Sf3 tan2 e = 0. From the f"rrst equation and by observing the curve in Fig. 17, we can readily see that the slope
of the curve is truly zero at 6 = 0. However, we must investigate the other factor. Setting it equal to
zero, we have
1- Sf3 tan2 6 = 0
or
tan2 e = 3fs
or
tan 6 = ±...(ii; = ±0.7746
Taking the positive value for the first quadrant, we find in the tables that
(J = 37.8°
Note that the negative value of tan 6 gives the angle 142.2°. The tangent line is also horizontal
at this point. You can see that this angle is about 142.2° by drawing on the graph of Fig. 17 a hori
zontal line tangent to the curve in the f"rrst and second quadrants and reading off the angle 6 at the
point of tangency. You can see from the graph that the tangent is also horizontal when 8 = 270°. However, we did not
obtain this value in our result because tan 8 was used in the calculations and tan 270° is not defined. Hence,
the answer for c) has four parts: 0°, 37.8°, 142.2°, and 270°. This example shows how much of a background in trigonometric equations is required to solve prob
lems involving polar equations. It also demonstrates the importance of using a figure to check the work.
Example3
lf the angle 8 in example 2 is increasing at a rate of 2 radians per second, at what rate is the radius vector increasing when 8 = 45°?
Solution Since the example calls for a. time rate of change of r, we need a derivative of r with respect to
t . F 1 2 h 3 · 3 Th d · · · h · · dr de 1m e. rom ex amp e , we ave r = sm 6. e envat1ve w1t respect to trrne IS- = 9 cos 38-dt dt"
49
-12Th fh d8 ·· 2 When 8 = 45°, 38 = 135° and so cos 38 = - - 2-. e angle's rate o c ange, dt' IS g1ven as .
dr Since dr is negative, r is decreasing.
dr d8 -= 9 cos 38-dt dt
An<
= 9 (-~ x 2 = -9J2or -12.7 units per second 2
Self-Test 9 1. Given the curve whose equation is r =a sin 30, find a) the angle which the curve makes with the initial line when 0 = 60°; b) the value of r when 0 = 60°; c) how fast the radius vector is increasing when 0 = 60° if the angle 0 is increasing at a rate of 72 radian per second.
o 2. Given the curve described by the equation, r =-,find a) the angle that the tangent to
1f
the curve makes with the radius vector when 0 = Jhrr; b) how fast the radius vector is increasing if the angle 0 is increasing at a rate of 2rr radians per minute. 3. Given the curve that is described by the equation r = 1.5 (1 +cosO), find a) the angle that the tangent line makes with the radius vector when 0 = 30°; b) the angle that the tangent line makes with the initial line when 0 = 30°; c) how fast the radius vector is increasing when the angle 0 is increasing at a rate of 2.0 radians per second.
41. Angular Velocity
In Art. 9 we discussed circular measure and found that the length of the arcs, subtended by the central angle 0, of a circle having a radius r is expressed by the formula
s = rO
When the angle 0 is increased, the arc length will be proportionately increased also. If 0 is a function of time, so wills be a function of time. We may, therefore, take the derivative of s = rO with respect to time and get
ds dO -=r-dt dt
Of course, r is a constant because the figure is a circle.
Now, ds is called the linear speed or velocity v of a point on the rotating radius at a dt
distance r from the center. Velocity v is also the measure of the rate at which s is described.
The rate of change of the angle 0 with respect to time is called the angular velocity of the radius vector. Angular velocity is commonly denoted by the Greek letter w (omega). Thus, by definition, we have
dO w=-dt
Substituting this into the preceding equation gives
ds dO v = dt = r dt = rw
which is a very useful relation between linear velocity and angular velocity as applied to a wheel or a circular path.
50
If 8 is in radians and t is in seconds, then w is in radians per second: When it is desired to have the angular velocity expressed in revolutions per second, we simply divide w
by 21T, since one revolution is equivalent to 21T radians. A popular symbol for angular velocity in revolutions per second is n. Thus, by definition again, we have
or
Example
w = 21Tn
w n=-21T
A wheel with radius 7.5 revolves with an angular velocity of 20 revolutions per second. What is the a) angular velocity, in radians per second, and b) linear velocity of a point on the rim of this wheel?
Solution
a) Since one revolution is equivalent to 2 rr radians, 20 revolutions per second is equivalent to 40rr radians per second, the required angular velocity. Ans.
b) Applying v = rw, in which r = 7.5, gives v = 7.5 (40Jrr) = 300rrunits per second. Ans.
42. Angular Acceleration
As you just learned, for circular motion, linear velocity is the distance traveled by a point on the circle per unit of time, and angular velocity is the amount of increase in the central angle per unit of time.
Angular acceleration is related to angular velocity just as linear acceleration is related to linear velocity. Specifically, acceleration is the derivative of velocity with respect to time. The symbol for angular acceleration is a (Greek letter alpha), and it is defined by the relation
dw d 2 8 a=-=--
dt dt 2
Since, for circular motion, v=rw
then the linear tangential acceleration
dv dw at=- =r- =ra
dt dt
This establishes a useful relation between tangential acceleration and angular acceleration, as applied to a wheel or to any circular motion.
Example _ A wheel with radius 12 revolves so that its angular velocity is given by the equation w = 16 + 612, where
1 is in seconds and w is in radians per second. Find a) the angular acceleration when 1 = 4 second~; b) the linear tangential acceleration of a point on the rim when 1 = 4 seconds; c) the linear velocity of a point
on the rim as a function of time; d) the number of revolutions the wheel will make between 1 = 0 and 1 = 4 seconds.
Solution dw d 2
a) Angular acceleration is a= dt = dt (16 + 6t ) = 12t
When 1 = 4, the angular acceleration o: = 48 radians per second per second. Ans. b) The tangential acceleration is a 1 = ra Since r = 12, then a 1 = 576 units per second per second. Ans. c) The linear velocity is v = rw, or since r = 12,
v = 192 + 721 2 units per sec. Ans.
51
dO d) By defmition, w = ar· This must be written in differential form so that we can integrate to
fmci 0. Thus we get dO= wdt, which becomes f dO= f wdt = f (16 + 6t2) dt
or 0=16t+2t3 +C
When t = 0, 0 = 0, and thus C = 0. When t = 4, 0 = 16(4) + 2(64) = 64 + 128 = 192 radians. But 192
radians is equivalent to 1;~, or 30.56 rev. Ans.
43. Rectangular Components of Velocity
Let's consider a particle that traverses a circle at a· uniform rate of n rev. per sec
(revolutions per second.) Let P(x,y) be the position of the particle given in rec
tangular coordinates, Fig. 18. Let OH be the projection of OP on OX, and OJ be
y
X
Fig. 18. Rectangular Projections of Point P on Circle
the projection of OP on OY. Drawing the lines HP, /P, and OP gives the right tri
angle from which we may write the following relationships.
x = OH=r cos 0
and y = 0/ =PH= r sin 0
where r is the radius of the circle. We are given as the angular velocity of OP n rev per sec or 2nn radians per sec. In
formula form this is written
from which we obtain
Integrating this, we get
dO w=-=2nn
dt
dO= 2nn dt
0 = 2nn t+ C
If we consider that when t = 0, the particle is on OX, then C = 0.
Thus 0 = 2nn t = wt
Therefore, x = r cos 0 = r cos 2n n t = r cos w t
52
and y =r sin 8 =r sin 21rn t= r sin wt
The component of velocity parallel to OX is
V=dx X dt
Now, since we have x as a function of8 and as a function of wt, let us use both of these
functions to gain experience and show that we can arrive at the same answer by using
either.
Thus
But
x = rcos 8
dx d8 - =r(-sin8)dt dt
d8 -=w dt
or
or
and
x = r cos wt
dx . d dt = r(- sm wt) dt (wt)
d - (wt)=w dt
Substituting win both equations, we get
V dx . . 8 =- =-rw sm wt=-rw sm X dt
Since 8 = wt, either form is satisfactory. How~ver, the form with wt is more frequently
used. Note that Vx is negative in the first quadrant, which means that x is decreasing
when 8 is increasing.
or
The component of velocity parallel to OYis Vy = dy. Sincey = r sin wt, dt
dy d - = r(cos wt)- (wt) dt dt
Vy = rw cos wt = rw cos 8
Having found the rectangular components of velocity, Vx and Vy, we can find the
resultant velocity by the right-triangle relationship
V=..JVx2 + V/
Substituting the values previously found for Vx and Vy, we get
V= ..J(- rw sin wt)2 + (rw cos wt)2
= ..J(rw)2 (sin2 wt + cos2 wt)
But (sin2 wt + cos2 wt) = 1
Therefore, V=rw
which is exactly what we found in Art. 41.
53
Example . A flywheel 4 units in diameter is rotating at a speed of 10 rev. per sec (revolutions per second). At a
given instant, point P is on the rim and I unit above the level of the center of the wheel, as shown in Fig. 19.
Find a) the velocity of P; b) the x-component of the velocity of P; c) they-component of the velocity of P.
Fig. 19. Point P on Rotating Flywheel
Solution a) The angular velocity of a spinning radius of the wheel is
w= 27rn
Here n = 10 rev per sec. Thus
w = 2 7r ( 1 0) = 20 7r radians per sec From V= rw,
V = 2(2071") = 407r = 125.7 units per sec. Ans.
b) In Fig. 19, PH= 1 and it is perpendicular to OX. Then triangle OPH is a right triangle, from
which we may write
sinO=%
and obtain the angle
e = sin-1 (%) = 30°
Using Vx = -rw sin Wt = -rw sin 8, and substituting for r, w, and 8, we find the horizontal compo
nent of velocity to be
Vx =- 2 (207r) sin 30°
= -201T units per sec
= -62.8 units per sec. Ans.
c) In a similar fashion, obtain the vertical or y-component of velocity.
Vy=rwcosO
= 2(2071") cos 30° = 40rr(f)units per sec.
= 108.8 units per sec. Ans.
Self-Test 10
1. A point on the rim of a flywheel with a radius of 5 is 4 units above the level ofthe center of the wheel and has a horizontal component of velocity of- 120 units per sec. Find a) the vertical component of velocity for the given conditions; b) the angular speed, in revolutions per second.
2. The coordinates of a particle moving in a circle are x = 10 cos 4t andy= 10 sin 4t.
Find a) the x-component of velocity; b) the y-component of velocity; c) the resultant velocity; d) the x-component of acceleration; e) they-component of acceleration;[) the resultant acceleration;g) the angular velocity.
54
44. Acceleration While at Constant Speed An important concept in physics is the rate at which the tangent line to a cir
cularly revolving body changes. For example, a record on a phonograph may be revolving at a constant speed. A point on the edge of this record will be moving at the same constant speed. However, remember that velocity is directed speed. The direction at the point is defined as the direction of the line tangent to the circle (or the record) at that point. Thus, the velocity of the point is constantly changing in direction.
In an earlier text we defined acceleration as the second derivative of directed displacement with respect to time or the first derivativ~ of velocity with respect to tiine. In mathematical symbols this is written for the x-component as
d2x dVx a =-=--
x dt2 dt and for they-component as
d2y dVy ay = dt2 =dt
For circular motion, we found in Art. 43 that Vx = -rw sin wt and Vy = rw cos wt
Thus the x-component of acceleration of a body in circular motion is
a =!!_ lv) =-rw2 coswt X dt \ X
for constant angular speed. Similarly, they-component of acceleration is
ay = :t (v.v) = -rw2 sin wt
also for constant angular speed.
or
The resultant centrifugal acceleration is
ac=Vai+a~
ac = V(-rw2 COS wt)2 + (-rw 2 sin wt)2
= ~2)2 (cos2 wt + sin2 wt)
=-rw2
Centrifugal acceleration is the rate at which velocity increases in the direction away from a line through the center of the circle. Centripetal acceleration is the rate at which velocity increases in the opposite direction.
From the above calculations, the centrifugal acceleration of the body rotating at a constant speed is negative. This means that the acceleration is centripetal acceleration.
This relationship is very interesting indeed. We see that the acceleration is constant in magnitude because both the radius and the angular speed are constants. The negative sign is significant because it means that the acceleration is directed along the radius towards the center of the circle. This offers an explanation why a balancing force away from the center must be applied on a revolving body to keep it going in a circular path.
45. Circular Motion with Constant Angular Acceleration
Many problems lead us to situations where the angular velocity is not constant but changing. In Art. 42 we learned that the angular acceleration is related to angular
55
velocity by the equation.
dw -=a dt
which may be written as
dw= adt
If a is a function of time, we must find this before integrating; however, if a is a con
stant, we can readily integrate and get
w=at+C1
If, at timet= 0, w = k, then the angular velocity at any timet is
w =at+ k
From the definition of angular velocity, we have
or
dO w=-=at+k
dt
dO=(at+k)dt
which, when integrated, gives
at2 o =-+kt+C2
2
If we begin counting 0 when t = 0, then c2 = 0, and the equation for angular dis
placement becomes
at2 o = -+ kt
2
The equations for the position of a particle on a circle may now be written as
x = r cos 8 = r cos(~ a t 2 + kt)
y = r sin 8 = r sin (~a t2 + kt)
From these we get the equation of the velocity components
dO Vx = r (-sin 8) dt = -r (at+ k) sin 8
= -rw sin 8
dO Vy =r(cos8)-=r(at+k) cosO
dt
=rw cos 8
We can see that the preceding equations for the velocity components are similar to those
of Art. 43. The exception is that here w is not a.constant but a function of time.
By taking the time derivative of Vx, we obtain the x-component of acceleration,
56
which is
d (dw) . d8 a =- V = -r - sm 8 - rw (cos 8)-" dt " dt dt
B dw d d8 hi h h b · d · th din · · ut -d =a an -d = w, w c w en su stitute m e prece g equation gives us t t .
ax = -r a sin 8 - rw2 cos 8
= -r a sin w t- rw2 cos w t
Similarly, we get the equation for they-component of acceleration; thus
ay = :r Vy = r a cos 8- rw2 sin 8
=ra cos w t- rw2 sin w t
Note that these equations have components of ra, the tangential acceleration, mentioned
in Art. 42, as well as rw2 the centrifugal acceleration mentioned in Art. 44.
Let us go a step further and fmd the resultant of the components. Thus from
aR =.Jai +a~
we get
and
a;= (ra)2 cos2 8 -2 (ra) (rw 2 ) sin 8 cos 8 + (rw2 )2 sin2 8
Adding these two equations we get
ai +a; = (ra)2 (sin2 8 + cos2 8) + (rw2 ) 2 (sin2 8 + cos2 8)
Again we recognize that (sin2 8 + cos2 8) = 1, and so we get
ai + a; = (r a)2 + (rw2 ) 2 .
and the resultant acceleration
aR = ../(ra)2 + (rw2 ) 2
which shows us that ther.e is a right-triangle relationship between the components, such
as shown in Fig. 20.
Fig. 20. Radial and Tangential Components of Acceleration
57
The component ra. is usually called the tangential component of acceleration. It is
possible for a. to be negative, and thus for the component ra. to be directed in the op
posite direction.
Example
A particle moves on a circle with a diameter of 8, according to the laws= 13 + 2t2, where sis the dis
tance measured along the arc of the circle from a fixed point on the path to the moving point, and t is the
time, in seconds. When t = 2 sec, determine the following: a) the linear velocity of the particle; b) the an
gular velocity of the radius to the particle; c) the tangent.ial component of acceleration; d) the angular
acceleration of the radius to the particle; e) the magnitude of the resultant linear acceleration of the particle.
Solution a) Since s is given as a function of time, we may find
V = ds = 3t2 + 4t dt
Thus when t = 2 sec, V= 3(4) + 4(2) = 20 units per sec. Ans.
b) s· f . I . 8 d ds d 8 h I I . . V 20 5 d. mce, orc1rcuarmotwn,s=r ,an di=r dt=rw,t eanguarveoc1tylsW=;-=4= ra 1ans
per sec. Ans.
d2s c) The tangential component of acceleration at= ra.=2.
dt When t = 2, a1 = 6(2) + 4 = 16 units per sec2 (second squared).
ds 2 d2s Since- = 3t + 4t, 2 = 6t + 4.
dt dt
at 16 2 d) The angula: acceleration a.= --; = 4 = 4 radians per sec Ans.
Let us solve for the angular acceleration in another way.
v 3t2 + 4t
F . d fi . . dw rom 1ts e m1t1on, a.= dt and
w = -; = --4- , from which we get
and when t = 2,
dw 6t+4 -;u=-4-
dw 12+4 2 a.= -- = --- = 4 radians per sec
dt 4
This agrees with the original solution. e) The centrifugal acceleration
ac = -rw2 = -4(5)2 = -100 units per sec2
Thus the resultant of the two components has the value
aR = Y(16)2 + (-100)2 = V10,256 = 101.3 units per sec2. Ans.
Self-Test 11
1. A line rotates in a vertical plane according to the law 8 = 4t3 - 12t2 , where 8 gives the
angular position of the line measured in radians and t is the time, in seconds. If the
radius of the circle is 2 and t = 3 sec, determine the following: a) the angular
velocity; b) the linear velocity; c) the angular acceleration; d) the tangential accel
eration. 2. The velocity of a particle moving along a circular path varies according to the law
V = 2t3 - 6, where Vis the magnitude of velocity, in feet per second, and t is the time,
in seconds. If the radius of the circle is 2 and t = 2 sec, find the following: a) the
58
angular velocity of the radius; b) the angular acceleration of the radius; c) the tangential component of acceleration; cl) the magnitude of the resultant acceleration.
The Derivatives of Hyperbolic Functions
46. Definitions of Hyperbolic Functions
Since hyperbolic functions are used in solving certain engineering problems, we
should know what they are and how to use them. The name "hyperbolic functions" is
given to certain combinations of the exponentials e" and e-u. These combinations are
~(e" + e-") and ~(e" - e-"). They occur with sufficient frequency to merit special
names. The first combination is called the "hyperbolic cosine of u" and is defined as
cosh u = lh(e" + e-")
Note that the letter "h" identifies the function as a hyperbolic function, distinguishing it
from the trigonometric function cos u. The second combination is called the "hyperbolic sine of u," and it is defined as
sinh u = lh(e"'- e·")
The main reason for having the names of the hyperbolic functions resemble the names of the trigonometric functions is that the hyperbolic functions have many properties that
are analogous to the trigonometric functions. For example, the trigonometric functions such as cos u and sin u are easily identified
with the point (x,y) on the unit circle having the equationx2 + y 2 = 1. This is why they
are often called circular functions. By properly defining u, we may take x = cos u andy =
sin u. When we substitute for x andy in x 2 + y 2 = 1, we get the well-known identity
cos2 u + sin2 u = 1
In a similar fashion, the hyperbolic function cosh u and sinh u may be identified
with the point (x,y), but on a "unit hyperbola,"
x2- y2 = 1
If we letx =cosh u andy= sinh u,
then
and y 2 = sinh2 u = 14(e"- e-")2 = 74(e 2"- 2+ e-2")
When we subtracty2 fromx 2 , we get
x2- y2 = 74(e2u + 2 + e-2u)- 14(e2u- 2 + e-2u) = 74(4) = 1
which proves one of the basic relationships of hyperbolic functions, namely, that
cosh2 u - sinh2 u = 1
Thus The hyperbolic tangent is defmed similarly to that of the trigonometric function.
sinh u e" - e-u tanhu=--=--
cosh u e" + e-u
59
The remaining hyperbolic functions are defined in terms of cosh u and sinh u as follows:
cosh u eu + e-u cothu=-- =---
Some useful identities are
sinh u eu - e-u
1 2 sechu=--=--
cosh u eu + e-u
1 2 cschu =-- =--
sinh u eu - e-u
cosh u + sinh u = eu
cosh u- sinh u = e-u
1- tanh2 u = sech2 u
coth2 u- 1 = csch2 u
From the first two identities it should be apparent that any combination of the expo
nentials eu and e-u can be replaced by a combination of sinh u and cosh u, and vice versa.
47. Discussion of Hyperbolic Functions
Hyperbolic functions are of such importance and utility that their numerical values
have been calculated and tabulated at least for sinh u, cosh u, and tanh u. The values of
the other functions can be readily expressed in terms of these, so it is unnecessary to tab
ulate all of them. Some slide rules also have sinh u and cosh u scales, and the values of
these functions may be· read directly. Certain major differences between the hyperbolic and circular functions should be
noted. The first difference is that the circular functions are periodic while the hyperbolic
functions are not periodic. The second great difference is that the two functions differ
in the range of values that they can assume; for example:
sin x varies between - 1 and + 1
sinh x varies from- oo to +oo
cos x varies between - 1 and+ 1
cosh x varies between + 1 to +oo
tan x varies from- oo to +oo
tanh x varies from - 1 to + 1
We shall conclude this article with certain formulas which you will find very useful.
sinh (x + y) = sinh x cosh y + cosh x sinh y
cosh(x + y) = cosh x cosh y + sinh x sinh y
sinh 2x = 2 sinh x cosh x
cosh 2x = cosh2 x + sinh2 x
cosh 2x + 1 = 2 cosh 2 x
cosh2x- 1 = 2 sinh2 x
Thus from the preceding you can see that practically all of the circular trigonometric
identities have hyperbolic analogies.
60
48. Derivative of Hyperbolic Sine Function
Lety =sinh u, where u is a differentiable function of x. Then
dy = ..!!_ sinh u = -~J.!.. (eu - e-u)] dx dx dx [2
Now we know the derivatives of exponential functions. Thus
Hence
Therefore,
d du -(eu)=eu-dx dx
and d ( -u) _ -u du - e --e -dx dx
d1 _ 1 du du - -(eu- e u) = -(eu + e-u)- =cosh udx2 2 dx dx
d du -sinh u =cosh udx. dx
Add this to your list of equations.
49. Derivative of Hyperbolic Cosine Function
Lety =cosh u, where u is a differentiable function of x. Then
dy d d [1 - 1 dx = dx (cosh u) = dx 2 (eu + e u)J
_ 1 ( u -u) du -- e - e -2 dx
Therefore, d du
dx (cosh u) = sinh u dx
SO. Derivatives of Other Hyperbolic Functions
(40)
{41)
Because the other hyperbolic functions of u are defined in terms of sinh u and cosh u, the formulas for their derivatives can be determined from formulas 40 and 41 along with the rule for calculating the derivative of a quotient. The formulas for the derivatives of the remaining hyperbolic functions, are as follows:
d du dx (tanh u) = sech2 u dx (42)
d du - (coth u) =- csch2 u-dx dx
(43)
d du dx (sech u) =- sech u tanh u dx {44)
d du dx (csch u) =- csch u coth u dx (45)
Observe that aside from the pattern of algebraic signs, these formulas are the exact analogs of the formulas for the corresponding circular functions.
Example 1
Giveny = sinh(3x2), fmd :.
61
Solution 2 du
In this problem u = 3x ; thus dx = 6x. Hence
dy 2 2 dx =cosh (3x ) (6x) = 6x cosh (3x ). Ans.
Example 2 dy
Giveny =tanh (1 +x 2), find dx'
Solution
dy = sech 2 (1 +x 2)~(1 +x 2) dx dx
=2xsech2 (l+x2). Ans.
Example 3 2 dy
Given cosh 3y = tan (3x ), find dx.
Solution d d
dx (cosh 3y) = dx (tan 3x2)
d d sinh 3y dx (3y) = sec2 (3x 2) dx (3x 2)
dy sec2 (3x 2) (6x) 3- = _ __;_ _ __;__ dx sinh 3y
dy 2x sec2 (3x2) -= dx sinh 3y
Ans.
Self-Test 12
Find : in each of the following:
1. y =sinh 2x 5. y = csch ( 1) 2. y = cosh2 (tanx) 6. y = Ln (cosh 3x)
3. y = tanh 2x 3 7. y = Y,. sinh 2x- ~x
4. y = cosh2 5x- sinh2 5x 8. coth 2y = tan 3x
51. Derivative of Inverse Hyperbolic Sine Function
Since we discussed the derivatives of the hyperbolic functions, we should be curious
about their inverse functions. The derivatives of inverse hyperbolic functions are similar to those of the circular functions with some differences in signs.
Let us follow through the familiar steps for finding the derivative of an inverse hyperbolic sine of u written as sinh - 1 u.
Let y = sinh - 1 u, where u is a differentiable function of x. Then, from the definition of an inverse function, we get
u = sinhy
which we can differentiate with respect to x and get
du dy - =coshydx dx
62
or dy du -=---dx coshy dx
Now from cosh2 y- sinh2 y = 1, the fundamental relationship of hyperbolic functions,
we have
But
Hence
or
Substituting this into
we get
Therefore,
cosh2 y = 1 + sinh2 y
sinhy = u or
coshy=~
dy du =---
dx coshy dx
dy 1 du
dx=~ dx
d 1 du -(sinh- 1 u)= -dx ~dx
(46)
d 1 du Compare this with- (sin- 1 u) = y'l7 (equation 25), and note the sign difference.
dx 1-u dx
52. Derivatives of Other Inverse Hyperbolic Functions
We may readily establish the derivatives of the other hyperbolic functions in a
manner similar to that used for establishing the derivative sinh- 1 u. These formulas are
listed here for ready reference.
d 1 du - (cosh- 1 u) = ---=== dx Vu2=1 dx
d 1 du - (tanh- 1 u) = -- -dx 1- u2 dx
(The absolute value of u must be less than 1.)
d 1 du - (coth- 1 u) = --dx 1- u 2 dx
(The. absolute value of u must be greater than 1.)
!!:._ (sech- 1 u) =- 1 du dx uvl-u2 dx
d _ 1 du - ( csch 1 u) = - -dx u~dx
(47)
(48)
(49)
(50)
(51)
The chief merit of the preceding inverse hyperbolic functions lies in their usefulness
in integration, which we shall study in Part 4 of this series.
63
Example 1 _ dy Given y = sinh 1 (2x), find dx·
Solution
Apply formula 46. du
Let u = 2x, and thus dx = 2. Therefore,
dy- 2 dx- VI+ 4x2
Ans.
Example 2 -1 dy
Given y = coth (sec x), find dx·
Solution
Apply formula 49. du
Here u =(sec x); thus dx =sec x tan x. Hence,
dy 1 sec x tan x sec x dx = 1- sec2 x (sec x tan x) = - tan2 x - --ta_n_x
sec x cosx sin x =- esc x. Ans.
Self-Test 13
Find: in each of the following:
l. y = sinh- 1 (3x) 6. y = sinh- 1 O~x)
2. y = tanh- 1 (sinx) 7. y = cosh- 1 (~)
3. y=coth- 1 e) 4. y = sech- 1 (cosx) 9. sinh- 1 (3x) = sinh (3y)
5. y = csch- 1 (x 2 ) 10. y = 2 tanh- 1 (tan ~x)
64
1. Formula (6)
y=bx 3
dy = 3bx3- 1 = 3bx2 Ans. dx
3. Formula (6)
y = 16 (x 3 + 2x)4 = 16u4
dy =4 X 16u4 - 1 = 64u 3
du
u =x3 + 2x
du 2 -= 3x + 2 dx
Appendix
Solutions to Self-Tests
Self-Test 1
2. Formula (6)
y = b (x + 2)3 = bu 3
u=x + 2
~= 1 dx
..!!!._ = 3bu 2 X du = 3b(x + 2)2 Ans. dx dx
dy = 64u 3 X du = 64 (x 3 + 2x)3 (3x2 + 2) Ans. dx dx
4. Formula (4)
y= (x + b)n (x 2 -c)m =u X v
u = (x + b)n
du n-1 -=n(x +b) dx v = (;2 -c)m
dv 2 m-1 - = 2mx(x -c) dx
dy dv du n 2 m-1 2 m n-1 -=u-+v-=(x+b) X2mx(x -c) +(x -c) Xn(x+b) dx dx dx
= 2mx (x + b)n (x2 -cr-1 + n(x2 -c)m (x + b)n-t Ans.
65
5. Formula (3)
y = (3x2 + 2x - 3)2 + Jx 2 + 12 = u + v
u = (3x2 + 2x- 3)2
du 2 - =2(3x + 2x- 3)(6x + 2) dx
v = (x 2 + 12)112
dv 1 2 -112 x d- = -2 (x + 12) (2x) = _2 ___ 11_2
x (x + 12)
X
dy du dv 2 x -=- +-= 2(3x + 2x- 3)(6x + 2) + Ans. dxdxdx ~
6. Formula (4)
7.
y = (x 2 - 1)2 (x2 + 2)3 = u X v
u = (x2 - 1)2
du 2 - =4x(x -1) dx
v = (x 2 + 2) 3
dv 2 2 - = 6x (x + 2) dx
dy dv du 2 2 2 2 2 3 2 - =u- + v -= (x -1) X 6x(x + 2) + (x + 2) X 4x(x - 1) dx dx dx
= 6x(x2 - l)(x2 - 1)(x2 + 2)2 + 4x(x2 + 2)(x2 - 1)(x2 + 2)2
= [6x(x2 - 1) + 4x(x2 + 2il (x2 -1)(x2 + 2)2
= 2x(3x2 - 3 + 2x2 + 4)(x2 -1)(x2 + 2)2
= 2x(5x2 + 1)(x2 - l)(x2 + 2)2 Ans.
Formula (5)
(3x + 1)2 u y=
(2x2 - 1)3 v
u = (3x + 1)2
du dx = 6(3x + 1)
v = (2x2-1)3
dv = 12x(2x2 -1)2 dx
66
du dv dy v~-u~ (2x2 -1)3 X6(3x+l)-(3x+l)2 Xl2x(2x2 -1)2
v2 (2x2 _ l)6
6(2x2 - 1) (2x2 - 1)2 (3x + 1) -12x(3x + 1) (3x + 1) (2x2 - 1)2
(2x2 - 1)6
= [12x2 - 6- 36x2 -12x] (2x2 -1)2(3x + 1) = -6(4x2 + 2x + 1)(3x + 1) Ans
(2x2- 1)6 (2x2 -1)4 .
8. Formula (4)
y =x(x2 + 4x -1)512 =u X v
u=x
du - =1 dx
2 5/2 v= (x + 4x -1)
dv = ~ (x2 + 4x- 1)312 (2x + 4) = S(x2 + 4x -1)312 (x + 2) dx 2
dy = u dv + v du = Sx(x + 2) (x2 + 4x- 1)312 + (x2 + 4x- 1)512 X 1 dx dx dx
= (Sx2 +lOx) (x2 + 4x -1f12 + (x2 + 4x -1) (x2 + 4x -1)312
2 2 2 3/2 2 2 3/2 = (Sx + lOx + x + 4x -1) (x + 4x- 1) = (6x + 14x- 1) (x + 4x- 1) Ans.
Self-Test 2
1. (x-y)y2 +x+y=O
( ) d(y)2 2 d(x-y) x-y ~+y ~
dx dy +- +- =0
dx dx
(x-y)2y dy + 'i-dy)y2 + 1 + dy = 0 dx \' dx dx
dy 2 dy 2 2 dy dy 2xy ~ - 2Y dx + y - Y dx + 1 + dx = 0
dy 2 dy dy 2 2xy - - 3y - + - =- y - 1
dx dx dx
dy 2 2 dx (-3y + 2xy + 1)=-y -1
dy -y2 -1
dx = -3y2 + 2xy + 1 3y2 -2xy-l Ans.
67
+ 2y-dy) dx
dy
Ans.
2 2 112 2y(x + y ) -y
1/2 112 3. {y+x) +(y-x) =a
~(y + x)-112 (ay + 0 + ~(y -x)-112(dy _) = 0
2 \d.x J 2 dx 1}
dy 1 1 dy X-+--+ ---X---- =0
.JY+X dx ...;y+x .jy _ x dx ~
(Jy:x + Jy~x) ix = ~
(.JY-X+.JY+X) dy = JY+X-~ .../Y+X ~ dx JY+X ,jy -x
~-~ -= Ans. dy
dx a
dy x dx- y
------- + = 0 x2
68
dy xy4 y dx = x2y3 = ; Ans.
5. (x + y)(x-y)2 =b2
(1 +ix) (x-y)2 +2(x+y)(x-y) (~-:)=o 2 2dy . dy
(x-y) + (x-y) dx +2(x+y)(x-y)-2(x+y)(x-y)dx=O
2 dy 2 [(x-y) -2(x+y)(x-y)) -=-(x-y) -2(x+y)(x-y)
dx
dy
dx
-(x-y)- 2 (x + y)
(x-y)- 2.(x + y)
-x + y - 2x - 2y 3x + y = -- Ans.
X - y - 2x - 2y X + 3y
3 2 2 dy 2dy 4x - 8xy - 8x y - + 3y - = 0
dx dx
2 2 dy 2 3 (3y - Sx y) - = 8xy - 4x
dx
dy
dx
7. 2x + xy + 3y = 6
8xy2 -4x3
3y2 - Sx2y
dy dy 2+y+x- + 3dx=O
dx
dy dy X-+ 3-=-y-2
dx dx
dy -:Y- 2 Ans. dx = ---;-+3
Ans.
69
d2y (x + 3)(- fx)+ (y + 2) X 1
dx 2 (x + 3)2
(x + 3)(y + 2) + y + 2 x+3
8. y = (x + y)
1 2
(x + 3)2
dy = ~(X+ y)_l/2 ~ + dy ) dx 2 \ dx
dy 1 1 +
dx
X dy =
X dy dx
dy
dx dx 2.Jx + y
2y + 4
(x + 3)2 Ans.
dy
dx --- Ans.
2v'X'+Y- 1 2y - 1
~ -2 ~ 1 ~ - 2 =-(2y-l) X 2- =- X 2-dx dx (2y- 1l dx
2 x---(2y- 1)2 (2y- 1)
dy dy 2x - 4 + 2y - + 6- = 0
dx dx
dy (2y + 6)- = 4 - 2x
dx
dy 4- 2x
dx = 2y + 6
Forx=5,y=1
dy 4-10 -6 -3 m=-= --=-=-
dx 2+6 8 4
(y-yl)=m(x-xl)
-3 (y- 1) = 4 (x- 5)
4y -4=-3x + 15
3x + 4y = 19· Ans.
70
2 Ans.
2dy 2dy 3y - = 8x-(x - + 2xy)
dx dx
2 2 dy (3y + X ) dx = 8x - 2xy
dy 8x-2xy
dx 3y2 +x2
For x =-2,y = 2
dy -16 + 8 m = dx = 12 + 4 2
(y -yl) = m(x -x 1)
1 (y-2)=-2(x+2)
2y -4 =-x-2
dx - = 6t dt
y = t2
dy - = 2t dt
x + 2y=2 Ans.
3. y=u6
dy = 6us du
u= 2 +x2
1=2xdx du
dx -=-du 2x
dy s 1 s dx = 6u + 2x = 12u x Ans.
Self-Test 3
dx 2 - = 3t dt
dy - = 2t dt
dy 2t 2 -=-=-AnSa dx 3t2 3t
4. y = u2 + 3u + 8
71
dy -=2u+3 du
u=2x+3
1 = 2dx du
dy 1 dx = (2u + 3) + 2 = 4u + 6 Ans.
S. y = 2V2 + 2V-2
dy -3 . 4. 40 - 4 dV= 4V-4V =4V- vl = vl
V = (3x + 2)213
1 = ~(3x + 2)-113 (. :v) dx (3x + 2) 113 -= dV 2
dy =(40 -4\. (3x + 2)113
dx v3/ 2
8(0 -1)
vl(3x + 2)113 Ans.
2/3 6. y = w
dy 2 -1/3 -=- w
dw 3
1 = 2x(J:)
dy 2 - 113 . 1 4x dx = 3w ..,. 2x·= 3w1/3 Ans.
Self-Tesf 4
1. y = cos6x=cos u
u =6x
du =6 dx
dy 6"6xA - =- s1n ns. dx
3. y = 3 tan 3x = 3 tan u
u = 3x
du = 3 dx
dy = 9 sec2 3x Ans. dx
S. y=sin2 3xcos2x=uXv
u = sin2 3x
du = 6 sin 3x cos 3x dx
2. x =4sin 2t=4 sinu
u = 2t
du =2 dt
dx - = 8 cos 2t Ans. dt
4. y = sin 2x cos 3x = u X v
72
u=sin2x
du -=2cos2x dx
v=cos3x
dv . - =-3sm3x dx
dy = 2 cos 2x cos 3x - 3 sin 2x sin 3~ Ans. dx
v =cos2x
dv . - =-2sm2x dx
dy = 6 sin 3x cos 3x cos 2x - 2 ·sin 2x sin2 3x Ans. dx
6. y = 3x2 - sin Sx
7.
dy - = 6x - S cos Sx Ans. dx
1 1 3 1 1 y=-sec2x+-tan 2x=-u+-v
2 6 2 6
u=sec2x
du -=2sec2xtan2x dx
v = tan3 2x
dv 2 2 -=6tan 2xsec 2x dx
dy 1 du 1 dv -=- -+--dx 2 dx 6 dx
= sec 2x tan 2x + tan2 2x sec2 2x Ans.
9. r 1-sin3 9
cos 9
u v
u = 1- sin3 9
du 3 . 2 -=- sm 9cos9 d9
v =cos 9
dv -=-sin 9 d9
dr -3 sin2 9 cos2 9 + sin 9 (1 - sin3 9) -= d9 cos2 9
-2 sin9 ·3 = -3 Sin 9 + -- (1- sm 9)
cos2 9
8. r=a(sin29-rcos29)
dr . - =a (2 cos 2 9-2 Sin 2 9) d9
= 2a (cos 2 9 -sin 2 9) Ans.
10. xy +cotxy=O
y + x :- (csc2 xy) ~ + x :) = 0
dy 2 dy 2 y + x dx - y esc xy - x dx esc xy = 0
dy 2 2 - (x-xcsc xy)=ycsc xy-y dx
dy -y(l - csc2 xy) y -= =-- Ans. dx x(1 -esc xy) x
= -3 sin2 9 +tan 9 sec 9 (1- sin3 9) Ans.
73
11. y=sinx
12.
1.
3.
dy --- = cosx dx
1r ..j2 a) cos 4" =-2- Ans.
. 2 X 180 114.59° = 114°35'24" b) 2 radians = 3_1416 =
cos 114°35'24" =-cos 65°24'36" =-0.4161 Ans.
tan 8 la I =I --=-- tan8
a tan 8a tan 8a
dl Ia 2 sec2 8 -=--sec 8=1 -- Ans. d8 tan8a atan8a
Self-Test 5
. -1 y =Sin 2x 2. y = tan-1 (2x-1)
dy 2 dy 2 -=
dx -J 1 - (2x)2
2 Ans.
-J1-4x2
2
4x 2 -4x + 2
y =COS-I 3x 4. y = sec-1 3x
dy -3 dy 3 - = Ans. -= dx
-J 1- 9x2 dx
3x-J9x2 -1
5. y = csc-1 (2x + 1) 6. y =coC1 (x2 -1)
dy 2 - =- --~:=:::;::== dx (2x + 1) ..J(2x + 1)2 - 1
2
(2x + 1) -J 4x2 + 4x
1 Ans.
dy -2x
dx 1 + ~2 -1)2
74
2 Ans. 2x-2x+1
Ans.
x-J9x2 -1
-2x Ans.
7. e = sin-1 G)
"9.
1 1 de 1 2 dx ;:I 1 r-;
-Y4-x2 2 4
1 Ans. .J 4 -x2
e=tan-{~~) x+1
x-1 u=--
X+ 1
du (x + 1)- (x- 1)
dx (x + 1)2
de X
2
(x + 1)2
2
1 -1 1 8. e = 3 tan 3x
Ans .
2
dx 1+c-1j
(x + 1)2 X
x 2 + 2x + 1 + x 2 - 2x + 1 (x + 1 )2
X + 1 (x + 1)2
(x + 1)2 2 2x2 + 2 X (x + 1)2 - x2 + 1 Ans.
10. y=xsin- 1 x+~
=sin -I x + __ x __ + x Ans.
~~
Self-Test 6
u = 2x -J
du _2 -2 -=-2x =dx x2
dy -2ex dx- T Ans.
3. y = (2x2 -2x + l)e2 x
dy (4 2X 2X 2 - = x - 2)e + 2e (2x - 2x + 1) dx
2. 2 -x y =e
dy -x2 - = -2xe = -2 xy Ans. dx
= e2x (4x- 2 + 2(2x2 - 2x + I)]= e2 x (4x- 2 + 4x2 - 4x + 2) = 4x 2e2 x Ans.
75
4. Y = (3 cos x + sin x)e3x
dy . 3X 3X · 3X -=(-3smx+cosx)e +3e (3cosx+smx)=10e cosx Ans. dx
7. exsiny=2 8.
.V dy ex+ e- -= 0
dx
X
dy = -e =-e(x-y) Ans. dx eY
sinx + cosx y=
X X dy e siny + (e cosy) dx = 0 dy (cosx- sin x)ex- ex(sin x +cos x)
dy -ex siny
dx ex cosy -tany Ans.
1. y =Ln (x 2 + 2x)
dy 2x + 2 Ans.
dx x2 + 2x
2 3. y =log-
X
u = 2x -1
du -2 -2 -=-2x dx x2
dy ·= loge ( -2) dx 2 2
- X X
= xloge(-2)= -loge ;! x2 x
Ans.
dx
-2ex sinx -2 sinx =
Self-Test 7
2. y = (Lnx) 3
dy = ~ (Lnx)2 Ans. dx X
4. y =Ln cosx
dy = - sin x = - tan x Ans. dx cosx
76
Ans.
S. y=Ln(tanx+secx)
dy sec2 x + secx tanx -= dx tan x + sec x
sec x (sec x + tan x) = = secx Ans.
secx + tanx
x2 7. y=Ln --2
1 +x x2
u=~
du 2x (l + x 2)- 2x (x2 ) -= dx (l +x2) 2
dy (1 + x 2) (2x) X
dx x2 (1 + x2)2
9. e<x + y) = Ln ::
2
(1 + dy)e<x + ~ = ~ (y -x ~) dx x y2
2x
(x + y) dy (x + y) 1 1 dy e +-e =----
dx x ydx
dy fe<x. + y) + ~) = ~- e (x + y) dx \ y x
dy (ye<x + y) + 1\ 1 -xe(x + y)
dx y J x
Ans.
dy y [1 -xe<x + y)J y -xy e<x + y)
dx x [ 1 + ye<x + Y>] x +xy e<x + y)
1 d 1 -_.!!. = 1 + - X 2x (x2 + 9) 2 dx 2
X =1+ q-;;
dy ( 1
dx= x+..Jx2+9
Ans.
dy 2 2 2x -= 2xLnx +x xdx x2
= 2x Ln x 2 + 2x = 2x(l + Ln x 2 ) An~.
Ans.
77
1
10. y =Ln (1 +x 2 ) 2 - x tan-1 x
dy
dx
11. xy = Ln (x + y)
X X
y + x dy = _1_ (1 + ddyx) dx x +y
( 1 ) dy 1 \x- x+y dx = x+y -y
(x2 +xy- 1\dy = 1-xy-y 2
X+ Y jdx X+ y
- (t.an-1 x + _x )= x 1 + x2 1 + x2
-1 X -tan x---
1 +x2
12. y=Ln(Lnx)
dx = (Ln ~)G)= x L~ x Ans.
dy 1-xy -y2
d 2 AnL X X + xy -1
13 . y= 3 secx
1.
Lny=Ln 3secx
Ln y = sec x Ln 3
1 dy --d = (Ln 3) tanx secx )' X
dy -= y Ln 3 tanx secx Ans. dx
Or, use formula 35 directly :
dy = 3sec x Ln 3 tan x sec x Ans. dx
e 00 10° 15° 20°
r 4 2$ 2Vl 2
e r
e 240° 250° 255° 260°
r 4 2.../3 2Vl 2
14. y =xLn x
Lny = LnxLnx
Lny = (Ln x)2
1 dy 1 -- = (2 Lnx)y dx x
dy 2y Lnx -= -- Ans. dx x
Self-Test 8
30° 90° 100° 105°
0 0 2 2.,fi
270° 330° 340° 345°
0 0 2 2.,fi
78
110°
2...[3
350° 360°
2...[3 4
3.
()
r
4.
I () 00 30° 45° 60° 90° 120° 131°48'38" 135° 150°
r 5 2 + 3y'3 2
3v'2 2+-2- 3.5 2 .o.s 0 2- 3v'2
2 2- 3y'3
2 or -0.12 or -0.60
180° 210° 225° 228°11 '22" 240° 270° 300° 315° 330° 360°
-1 2 _3v'3 2
2- 3..;2 2
or-0.60 or-0.12
() 00 15° 22°30' 30°
r 0 2 2v'2 20
(} 1120° 135° 150°
r l-2v'3 -4 -2v'3
() 240° 247°30' 255°
r 2v'3 2v'2 2
() 345° 360°
r -2 0
0
45°
4
157°30'
-20.
270°
0
0.5 2 3.5 2 + 3..;2 2
2 + 3vG 2
5
60° 67°30' 75° 90° 105° 112°30'
2../3 2v'2 2 0 -2 -2.J2
165° 180° 195° 202°30' 210° 225°
-2 0 2 2v'2 2$ 4
285° 292°30' 300° 315° 330° 337°30'
-2 -2v'2 -2J3 -4 -2v'3 -2J2
80
Self-Test 9
1. a) First find the angle that the curve makes with in radius vector, as in example 2 a). Remember
that 0 is the angle that the radius vector makes with the initial line.
r =asiu30 dr = 3a cos 30 dO
tan 1/1 = r dO = a sin 30 dO = ~ tan 30 dr 3a cos 30 dO 3
0 0 1 0 For = 60 , tan 1/1=- tan 180 = O,and 1/1 = 0°
3
1/J = 0 + 1/1 = 60° + 0° = 60° Ans.
b) r=asin30=asin180°=aXO=O Ans.
c) dr = 3a cos 30 dO
dr dO -=3acos30-dt dt
dr
dt 3a Cos 18oo X 1 3 -3a
2 = 2 a (-1) = 2 Ans.
81
2 .. a) ()
r=tr
1 dr =- dfJ
1r
() - dfJ
r dfJ 7r 7r tan 1/1 = -- = --= fJ =- = 1.5708
dr 1 2 - dfJ
1 b) dr = - dfJ
1r
1r
dr 1 dfJ 1 = -- = - x 27r = 2 Ans.
dt 1r dt 1r
3. a) r = 1.5(1 +cos fJ) = 1.5 + 1.5 cos()
dr = -1.5 sin() dfJ
tan 1/1 = _.!_!!!!__ = 1.5 (1 + cos fJ) dfJ dr -1.5 sin fJ dfJ
-(1 +cos fJ) -(1 + cos 30°) = -{1 + 0.86603) = -3.73206
sin() sin 30° 0.5
1/1 = 180° - 75° = 105° Ans.
b) ¢ = () + 1/1 = 30° + 105° = 135° Ans.
c) dr = -1.5 sin() dfJ
dr . dfJ . o - = -1.5 SID fJ- = -1.5 SID 30 X 2.0 = (-1.5) (0.5) (2.0) = -1.5 dt dt
The radius is decreasing at 1.5 radians per second. Ans.
Self-Test 10
. y 4 J ·2 3 1. r = 5, y = 4, SID fJ = - = - , COS (} = 1 - SID (} = -
r 5 5
Use the formula on page 53 to find w
V =-rwsinfJ X
4 -120=-5 X- W=--4W
5
W=30
82
2.
2.
3 a) Vy=rwcos8=S X 30X S = 90unitsfsec Ans.
b) Angular velocity = W = 2rr n
30 = 2rr n
30 IS n = -=-rev/sec Ans.
2rr rr
a) X =10cos4t
Vx = dx = --40 sin 4t Ans. dt
b) y = 10 sin 4t
Vy dy
Ans. =- = 40 cos 4t dt
c) v = Jvx2 + v 2 y Jc--40 sin 4t)2 + (40 cos 4t)2 )I600(sin2 4t + cos2 4t)
.JI600 = 40 Ans.
d) ax d V x d(--40 sin 4t)
= -- = =-160 cos 4t Ans. dt dt
e) ay d Vy d(40 cos 4t)
= -- = = -160 sin 4t Ans. dt dt
d8 d(4t) 0
g) w = - = -- = 4 radums per second Ans. dt dt
Self-Test 11
a) d8
w = - = 12t2 - 24t = 12 X 9 - 24 X 3 = 36 radians per sec Ans. dt
b) V = rX w = 2 X 36 = 72unitspersec Ans.
c) dw
a = - = 24t - 24 = 24 X 3 - 24 = 48 radians per sec2 Ans. dt
d) at = r X a = 2 X 48 = 96 units per sec2 Ans.
v 2t3 - 6 2 X 8-6 = 5 radians per sec Ans. a) w =- =
r r 2
b) dw 6t2 6X4
12 radians per sec2 Ans. a =-=- -2- = dt 2
83
c) at = r X a = 2 X 12 = 24 units per sec2 Ans.
d) ac = -rw2 =-2 X 25 = -50
aR = )at2 + ac2 = J242 + (-50)2 = .j3Qi6 = 55.46,or 55.5 units per sec2 Ans.
1. y = sinh 2x
dy - = 2 cosh 2x Ans. dx
3. y = tanh 2x 3
5. y = csch (~)
7.
- =-- csch - coth -dy 1 (X) (X) dx 2 2 2
1 . 1 y=-smh2x--x
4 2
dy 1 1 -=- cosh2x-dx 2 2
1 = 2 (cosh 2x-1) Ans.
1. y = sinh - 1 (3x)
dy 3 Ans. -=
dx
Ans.
Self-Test 12
2. y = cosh2 (tan x) = u2
u = cosh (tan x)
du 2 . - = sec x smh (tan x) dx
dy - = 2u = 2 cosh (tan x) du
dy22"() - = sec x smh tan x cosh (tan x) Ans. dx
4. y = cosh2 Sx - sinh2 Sx
dy = 0 An s. dx
6. y = Ln (cosh 3x)
dy = __ l_ X 3 sinh 3x = 3 tanh 3x Ans. dx cosh 3x
8. coth 2y = tan 3x
(-2 csch2 2y>(ix)= 3 sec2 3x
dy - = ----=-dx
3 sec2 3x Ans.
-2 csch2 2y
Self-Test 13
2. y = tanh - 1 (sin x) = tanh - 1 u
84
. du u = sm x, - = cos x
dx
dy X cosx -=
dx
cosx
cos2 x cosx secx Ans. =---
3. y = coth - 1 {~) = coth - 1 u
7.
9.
I -1 u=-=x
X
du _2 -1 - = -x =-dx x2
dy -1 -=----x--dx x2 - x2
x2
-1 = --- =--- Ans.
dy -2x -2 -=--.:::....--=----dx .r----:
x2v1 + x4
y = cosh-1 &'
dy ex -= dx
Je2x - 1 Ans.
sinh -l (3x) = sinh (3y)
3 dy 3 cosh (3y)-
dx J1 + 9x2
dy dx
J1 + 9x2 cosh (3y)
Ans.
4. y = sech-1 (cosx) = sech-1 u
Ans.
8.
85
du . u = cosx, dx = -smx
dy -sin x dx =- --=;:=:;;:::-
cos x .J1 - cos2 x
sinx = =--
COS X sin X COS X
X u=-2'
du 1 dx = 2'
dy -=----X dx
y = tanh-1 (4x2 )
2
Ans.
dy 8x Ans.
dx - 16x4
= secx Ans.
10.. y = 2 tanh-1 ~an~)= 2 tanh-1 u
X u= tan 2,
dy 2 -=
du dx
dx 2 X - tan -
2
X
1 2 X
2 sec 2
2 X sec
2 2
~"'X . 2
Sin 1 2
--x-+ 2 X cos2 2 cos
2
secx Ans. cosx
sec2 X
2
1 - tan 2 X
2
X) 2 - 2 X
cos 2-
86
( . ,x ) Sin -1 2
--+ 1---2 X X
cos 2 cos2 2
. 2 X Sin
2 cos (2 X D
EXAMINATION 6618C
Calculus: Function and Use, Part 3 When you feel confident that you have mastered the material in this study unit,
complete the following examination. Then submit only your answers to school
headquarters for grading, using one of the examination answer options de
scribed in your first shipment. Send your answers for this examination as soon
as you complete it. Do not wait until another examination is ready.
Questions 1-15: Select the one best answer to each question.
dy 1. Calculate dx when xy3 + y = 3x.
A.-3-y3 + I
B.-3-y3- I
dy (3x + 1)3 2. Calculate- when y2 = -----::--~
dx 9x + 2 ·
-3(3x + 1)2 A. -::--:::------:~
2y(9x + 2)2
-3(3x + 1)2 B.
y 2(9x + 2)2
c. 3 + y3
I + 3xy2
D. 3- y3
l + 3xy2
9(3x + 1)2 (6x + I) c. --~~-~~-
2y(9x + 2)2
9(3x + 1)2 (6x + I) D. ---::-=----::~-
y2(9x + 2)2
3. Given the functions y = 3x3 + 6x and x = 2t2 + t, find ~.
A. 9t2 + 6
B. (9x2 + 6)(4t + I)
4. Find dx when y = cosh (bx2).
A. 2bx cosh (bx2)
B. -2bx cosh (bx2)
S. Find : when y = Ln (sinh 2x).
A. 2 cosh 2x
B. 2 coth 2x
9x2 + 6 c. 4t + I 4t + I
D.---9x2 + 6
C. 2bx sinh (bx2)
D. -2bx sinh (bx2)
C. 2 sech 2x
D. 2 csch 2x
6. Find : when sinh 3y = cos 2x.
A. -2 sin 2x
B. -2 ~in 2x smh 3y
2 2x C. --tan(-)
3 3y
D. _ 2 sin 2x 3 cosh 3y
7. Find the derivative of y = cos (x2) with respect to x.
A. -sin (2x) C. - sin (2x) cos (x2)
B. -2x sin (x2) D. -2x sin (x2) cos (x2)
8. Find the derivative of y = sin2 (4x) cos (3x) with respect to x.
A. 8 sin (4x) cos (3x)- 3 sin2 (4x) sin (3x)
B. 8 cos (4x) cos (3x) - 3 sin2 (4x) sin (3x)
C. 8 sin (4x) cos (4x) cos (3x) - 3 sin2 (4x) sin (3x)
D. 8 sin (4x) cos (4x) cos (3x) - 3 sin2 (4x) sin (3x) cos (3x)
9. Find the derivative of y = ebx2 with respect to x.
A. bebx2 C. 2 bxebx2
B. bxebx2 D. bx2 ebx2
10. Calculate dy when y = (x2 + 2)e 4x as the derivative of a product, letting dx
u = x2 + 2 and v = e4x. Which of the following is a step in your solution?
du dv A.-= 2x + 2 C.-= 4xe3x
dx dx
B. du = 2x dx
dv D.-= 4xe 4x
dx
11. Which of the following represents : when y = e-2x sec (3x)?
A. 3e -2x sec (3x) tan (3x) - 2e -2x sec (3x) B. 3e -2x sec (3x) tan (3x) - 2xe -2x sec (3x)
C. 3e-2x sec (3x) tan (x) -2 e-2x sec (3x)
D. 3e-2x sec (3x) tan (x) - 2xe-2x sec (3x)
12. Calculate : if y = In (2x3 + 3x).
1 A. 2x3 + 3x
1 B. 6x2 + 3
2
2x3 + 3x C. 6x2 + 3
6x2 + 3 D.---
2x3 + 3x
dy ..!. 13. Calculate dx if Ln (x + y) = eY.
X X
eY xy+eY y2+y2 A.~~~~~--~
X X
e'Y xy+eYx2+y2
X X
eY xy+eY y2-y2 B. ..:.....,x:--=--.::.......,.x_;__..::~
eY xy-eY x2+y2
X X
eY xy+eY y2+y2 c. --..,--=----=~--=-~ !...
eY xy-eY x2+y2
X X
D. e Y xy + e Y y2 - y2 X !_
eY xy+eY x2+y2
14. Given the curve that is described by the equation r = 3 cos 9, find the angle that the tangent line makes with the radius vector when 9 = 120°.
A. 30° B. 45°
c. 60° D. 90°
15. A line rotates in a horizontal plane according to the equation 9 = 2t3- 6t, where (J is the angular position of the rotating line, in radians, and t is the time, in seconds. Determine the angular acceleration when t = 2 sec.
A. 6 radians per sec2 C. 18 radians per sec2 B. 12 radians per sec2 D. 24 radians per sec2
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