3.3 differentiation formulas in this section, we will learn: how to differentiate constant...
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3.3Differentiation Formulas
In this section, we will learn:
How to differentiate constant functions,
power functions, polynomials, and
exponential functions.
DERIVATIVES
The constant function: f(x) = c.
The graph of this function is the
horizontal line y = c, which has slope 0.
So, we must have f’(x) = 0.
CONSTANT FUNCTION
A formal proof—from the definition of
a derivative—is also easy.
0
0 0
( )'( ) lim
lim lim0 0
h
h h
f x h f xf x
hc c
h
CONSTANT FUNCTION
If n = 1, the graph of f(x) = x is the line
y = x, which has slope 1.
So,
You can also verify Equation 1 from the definition of a derivative.
( ) 1dx
dx
POWER FUNCTIONS Equation 1
Figure 3.3.2, p. 135
We have already investigated the cases
n = 2 and n = 3.
In fact, in Section 3.2, we found that:
2 3 2( ) 2 ( ) 3d dx x x x
dx dx
POWER FUNCTIONS Equation 2
For n = 4, we find the derivative of f(x) = x4
as follows:
4 4
0 0
4 3 2 2 3 4 4
0
3 2 2 3 4
0
3 2 2 3 3
0
( ) ( ) ( )'( ) lim lim
4 6 4lim
4 6 4lim
lim 4 6 4 4
h h
h
h
h
f x h f x x h xf x
h h
x x h x h xh h x
h
x h x h xh h
h
x x h xh h x
POWER FUNCTIONS
Comparing Equations 1, 2, and 3,
we see a pattern emerging.
It seems to be a reasonable guess that, when n is a positive integer, (d/dx)(xn) = nxn - 1.
This turns out to be true.
We prove it in two ways; the second proof uses the Binomial Theorem.
POWER FUNCTIONS
The formula
can be verified simply by multiplying
out the right-hand side (or by summing
the second factor as a geometric series).
1 2 2 1( )( )n n n n n nx a x a x x a xa a
Proof 1POWER RULE
If f(x) = xn, we can use Equation 5 in
Section 3.1 for f’(a) and the previous equation
to write:
1 2 2 1
1 2 2 1
1
( ) ( )'( ) lim lim
lim( )
n n
x a x a
n n n n
x a
n n n n
n
f x f a x af a
x a x a
x x a xa a
a a a aa a
na
POWER RULE Proof 1
In finding the derivative of x4, we had to
expand (x + h)4.
Here, we need to expand (x + h)n . To do so, we use the Binomial Theorem —as follows.
0 0
( ) ( ) ( )'( ) lim lim
n n
h h
f x h f x x h xf x
h h
Proof 2POWER RULE
This is because every term except the first has h as a factor and therefore approaches 0.
1 2 2 1
0
1 2 2 1
0
1 2 2 1 1
0
( 1)2
'( ) lim
( 1)2lim
( 1)lim
2
n n n n n n
h
n n n n
h
n n n n n
h
n nx nx h x h nxh h x
f xh
n nnx h x h nxh h
hn n
nx x h nxh h nx
POWER RULE Proof 2
a.If f(x) = x6, then f’(x) = 6x5
b.If y = x1000, then y’ = 1000x999
c.If y = t4, then
d. = 3r23( )
dr
dr
Example 1POWER RULE
34dy
tdt
If c is a constant and f is a differentiable
function, then
( ) ( )d dcf x c f x
dx dx
CONSTANT MULTIPLE RULE
p. 137
Let g(x) = cf(x).
Then,0
0
0
0
( ) ( )'( ) lim
( ) ( )lim
( ) ( )lim
( ) ( )lim (Law 3 of limits)
'( )
h
h
h
h
g x h g xg x
hcf x h cf x
hf x h f x
ch
f x h f xc
hcf x
ProofCONSTANT MULTIPLE RULE
4 4
3 3
a. (3 ) 3 ( )
3(4 ) 12
b. ( ) ( 1)
( 1) ( ) 1(1) 1
d dx x
dx dx
x x
d dx x
dx dxdx
dx
NEW DERIVATIVES FROM OLD Example 2
Let F(x) = f(x) + g(x). Then,
0
0
0
0 0
( ) ( )'( ) lim
( ) ( ) ( ) ( )lim
( ) ( ) ( ) ( )lim
( ) ( ) ( ) ( )lim lim (Law 1)
'( ) '( )
h
h
h
h h
F x h F xF x
hf x h g x h f x g x
hf x h f x g x h g x
h h
f x h f x g x h g x
h hf x g x
ProofSUM RULE
The Sum Rule can be extended to
the sum of any number of functions.
For instance, using this theorem twice, we get:
( ) ' ( ) '
( ) ' '
' ' '
f g h f g h
f g h
f g h
EXTENDED SUM RULE
If f and g are both differentiable,
then
( ) ( ) ( ) ( )d d df x g x f x g x
dx dx dx
DIFFERENCE RULE
8 5 4 3
8 5 4
3
7 4 3 2
7 4 3 2
( 12 4 10 6 5)
12 4
10 6 5
8 12 5 4 4 10 3 6 1 0
8 60 16 30 6
dx x x x x
dxd d dx x x
dx dx dxd d dx x
dx dx dx
x x x x
x x x x
NEW DERIVATIVES FROM OLD Example 3
Find the points on the curve
y = x4 - 6x2 + 4
where the tangent line is horizontal.
NEW DERIVATIVES FROM OLD Example 4
Horizontal tangents occur where
the derivative is zero.
We have:
Thus, dy/dx = 0 if x = 0 or x2 – 3 = 0, that is, x = ± .
4 2
3 2
( ) 6 ( ) (4)
4 12 0 4 ( 3)
dy d d dx x
dx dx dx dx
x x x x
Solution: Example 4
3
So, the given curve has horizontal tangents
when x = 0, , and - .
The corresponding points are (0, 4), ( , -5), and (- , -5).
Solution: Example 4
33
3 3
Figure 3.3.3, p. 139
The equation of motion of a particle is
s = 2t3 - 5t2 + 3t + 4, where s is measured
in centimeters and t in seconds.
Find the acceleration as a function of time.
What is the acceleration after 2 seconds?
NEW DERIVATIVES FROM OLD Example 5
The velocity and acceleration are:
The acceleration after 2s is: a(2) = 14 cm/s2
2( ) 6 10 3
( ) 12 10
dsv t t t
dtdv
a t tdt
Solution: Example 5
If f and g are both differentiable, then:
In words, the Product Rule says: The derivative of a product of two functions
is the first function times the derivative of the second function plus the second function times the derivative of the first function.
( ) ( ) ( ) ( ) ( ) ( ) d d df x g x f x g x g x f x
dx dx dx
THE PRODUCT RULE
Let F(x) = f(x)g(x).
Then,
Proof
0
0
( ) ( )'( ) lim
( ) ( ) ( ) ( )lim
h
h
F x h F xF x
hf x h g x h f x g x
h
THE PRODUCT RULE
0
0
0 0 0 0
'( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )lim
( ) ( ) ( ) ( )lim ( ) ( )
( ) ( ) ( ) ( )lim ( ) lim lim ( ) lim
( ) '( ) ( ) '( )
h
h
h h h h
F x
f x h g x h f x h g x f x h g x f x g x
h
g x h g x f x h f xf x h g x
h h
g x h g x f x h f xf x h g x
h h
f x g x g x f x
ProofTHE PRODUCT RULE
Find F’(x) if F(x) = (6x3)(7x4).
By the Product Rule, we have:
Example 6
3 4 4 3
3 3 4 2
6 6
6
'( ) (6 ) (7 ) (7 ) (6 )
(6 )(28 ) (7 )(18 )
168 126
294
d dF x x x x x
dx dx
x x x x
x x
x
THE PRODUCT RULE
If h(x) = xg(x) and it is known that
g(3) = 5 and g’(3) = 2, find h’(3).
Applying the Product Rule, we get:
Therefore,
Example 7
'( ) [ ( )] [ ( )] ( ) [ ]
'( ) ( )
d d dh x xg x x g x g x x
dx dx dxxg x g x
THE PRODUCT RULE
'(3) 3 '(3) (3) 3 2 5 11 h g g
If f and g are differentiable, then:
In words, the Quotient Rule says: The derivative of a quotient is the denominator times
the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
2
( ) ( ) ( ) ( )( )
( ) ( )
d dg x f x f x g xd f x dx dx
dx g x g x
THE QUOTIENT RULE
Let F(x) = f(x)/g(x).
Then,
THE QUOTIENT RULE Proof
0
0
0
( ) ( )'( ) lim
( ) ( )( ) ( )
lim
( ) ( ) ( ) ( )lim
( ) ( )
h
h
h
F x h F xF x
hf x h f xg x h g x
hf x h g x f x g x h
hg x h g x
We can separate f and g in that expression
by subtracting and adding the term f(x)g(x)
in the numerator:
Proof
0
0
'( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )lim
( ) ( )
( ) ( ) ( ) ( )( ) ( )
lim( ) ( )
h
h
F x
f x h g x f x g x f x g x f x g x h
hg x h g x
f x h f x g x h g xg x f x
h hg x h g x
THE QUOTIENT RULE
Again, g is continuous by Theorem 4 in Section 3.2.
Hence,
Proof
0 0 0 0
0 0
2
( ) ( ) ( ) ( )lim ( ) lim lim ( ) lim
lim ( ) lim ( )
( ) '( ) ( ) '( )
[ ( )]
h h h h
h h
f x h f x g x h g xg x f x
h hg x h g x
g x f x f x g x
g x
0lim ( ) ( )
hg x h g x
THE QUOTIENT RULE
The theorems of this section show
that:
Any polynomial is differentiable on .
Any rational function is differentiable on
its domain.
THE QUOTIENT RULE
Then,
THE QUOTIENT RULE Example 8
3 2 2 3
23
3 2 2
23
4 3 4 3 2
23
4 3 2
23
6 2 2 6'
6
6 2 1 2 3
6
2 12 6 3 3 6
6
2 6 12 6
6
d dx x x x x x
dx dxyx
x x x x x
x
x x x x x x
x
x x x x
x
The figure shows the graphs of the function of
Example 8 and its derivative. Notice that, when y grows rapidly (near -2),
y’ is large. When y grows
slowly, y’ is near 0.
Figures:
Figure 3.3.4, p. 141
Example 8
Don’t use the Quotient Rule every
time you see a quotient.
Sometimes, it’s easier to rewrite a quotient first to put it in a form that is simpler for the purpose of differentiation.
NOTE
For instance:
It is possible to differentiate the function
using the Quotient Rule. However, it is much easier to perform the division
first and write the function as
before differentiating.
23 2( )
x x
F xx
1 2( ) 3 2 F x x x
NOTE
GENERAL POWER FUNCTIONS Proof
2
1
2
11 2
2
1
(1) 1 ( )1( )
( )
0 1
n n
nn n
n n
n
nn n
n
n
d dx xd d dx dxx
dx dx x x
x nx
x
nxnx
x
nx
a. If y = 1/x, then
b.
GENERAL POWER FUNCTIONS Example 9
1 22
1( )
dy dx x
dx dx x
3 43
4
66 ( ) 6( 3)
18
d dt t
dt t dt
t
So far, we know that the Power Rule
holds if the exponent n is a positive or
negative integer.
If n = 0, then x0 = 1, which we know has a derivative of 0.
Thus, the Power Rule holds for any integer n.
POWER RULE – integer version
What if the exponent is a fraction?
In Example 3 in Section 3.2, we found that:
This can be written as:
1
2
dx
dx x
1 2 1 212
dx x
dx
FRACTIONS
This shows that the Power Rule is true
even when n = ½.
In fact, it also holds for any real number n,
as we will prove in Chapter 7.
FRACTIONS
a. If f(x) = xπ, then f ’(x) = πxπ-1.
b.
Example 10POWER RULE
3 2
2/3
(2 /3) 123
5/323
1
( )
Let
Then
yx
dy dx
dx dx
x
x
Differentiate the function
Here, a and b are constants. It is customary in mathematics to use letters near
the beginning of the alphabet to represent constants and letters near the end of the alphabet to represent variables.
PRODUCT RULE Example 11
( ) ( )f t t a bt
Using the Product Rule, we have:
PRODUCT RULE E. g. 11—Solution 1
1 21
2
'( ) ( ) ( )
( )
( ) ( 3 )
2 2
d df t t a bt a bt t
dt dt
t b a bt t
a bt a btb t
t t
If we first use the laws of exponents to rewrite
f(t), then we can proceed directly without using
the Product Rule.
This is equivalent to the answer in Solution 1.
1 2 3 2
1 2 1 2312 2
( )
'( )
f t a t bt t at bt
f t at bt
LAWS OF EXPONENTS E. g. 11—Solution 2
They also enables us to find normal
lines.
The normal line to a curve C at a point P is the line through P that is perpendicular to the tangent line at P.
In the study of optics, one needs to consider the angle between a light ray and the normal line to a lens.
NORMAL LINES
Find equations of the tangent line
and normal line to the curve
at the point (1, ½).
Example 12TANGENT AND NORMAL LINES
2/(1 ) y x x
According to the Quotient Rule,
we have:
Example 12Solution: derivative
2 2
2 2
2
2 2
2 2 2
2 2 2 2
(1 ) ( ) (1 )
(1 )
1(1 ) (2 )
2(1 )
(1 ) 4 1 3
2 (1 ) 2 (1 )
d dx x x xdy dx dx
dx x
x x xxx
x x x
x x x x
So, the slope of the tangent line at (1, ½) is:
We use the point-slope form to write an equation of the tangent line at (1, ½):
31 14 4 41 ( 1) or y x y x
Solution: tangent line Example 12
2
2 21
1 3 1 1
42 1(1 1 )
x
dy
dx
The slope of the normal line at (1, ½) is
the negative reciprocal of -¼, namely 4.
Thus, an equation of the normal line is:
712 24( 1) 4or y x y x
Solution: normal line Example 12
The curve and its tangent and normal
lines are graphed in the figure.
Figures: Example 12
© Thomson Higher Education
Figure 3.3.5, p. 144
At what points on the hyperbola xy = 12
is the tangent line parallel to the line
3x + y = 0?
Since xy = 12 can be written as y = 12/x, we have:
TANGENT LINE Example 13
1 22
1212 ( ) 12( )
dy dx x
dx dx x
Let the x-coordinate of one of the points
in question be a.
Then, the slope of the tangent line at that point is 12/a2.
This tangent line will be parallel to the line 3x + y = 0, or y = -3x, if it has the same slope, that is, -3.
Solution: Example 13
Equating slopes, we get:
Therefore, the required points are: (2, 6) and (-2, -6)
Solution: Example 13
22
123 4 2or or a a
a
The hyperbola and the tangents are
shown in the figure.
Figure: Example 13
© Thomson Higher Education
Figure 3.3.6, p. 144