chain rule,implicit differentiation and linear approximation and differentials
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1
Chain Rule
If y = f(u) is a differential function of u and u in turn is a differentiable function of x, that is u = g(x), then
y = f (g(x)) is a differentiable function of x and its derivative is given by the product
The chain rule states that the change in y w.r.t. x is the product of two rates of change.
dx
du
du
dy
dx
dy
2
Solution
• Differentiate g(x) =
• Write g(x) = where u = is the inner
function and is the outer function. Then
4
14
1
31u
x
x
x
x
31
4/1u
)(4
1)( '4
3' xuuxg
2
43
'
)31(
)3()1)(31(
314
1)(
x
xx
x
xxg
45
4
32
4
3
314
1
31
1
314
1
xxxx
x
x
x
31
4
3
Differentiation using Chain Rule
• y = f(u) and u = g(x)
dx
dunuu
dx
d nn 1
dx
duuu
dx
dcossin dx
duuu
dx
dsincos
dx
duuu
dx
d 2sectan dx
duuu
dx
d 2csccot
dx
duuuu
dx
dtansecsec
dx
duuuu
dx
dcotcsccsc
dx
duee
dx
d uu dx
du
uu
dx
d 1ln
4
Example
• Write the composite function in the form of f(g(x)) and then find the derivative using the chain rule.
• (i)
• (ii)
• (iii)
• (iv)
102 )1( xy
xy sin
)tan(sin xy
)cossin( xxy
5
Solution (i)
• (i)
•
• Let u = g(x) = 1 - and y = f(u) =
Then,
2x 10u
)2(10 9 xudx
du
du
dy
dx
dy
92 )1(20 xx
102 )1()())(( xufxgfy
6
Solution (ii)
• (ii)
• Let u = g(x) = sin x and y = f(u)
xufxgfy sin)())((
u
xudx
du
du
dy
dx
dycos
2
1 2/1
u
x
2
cos
x
x
sin2
cos
7
Solution (iv)
• (iv)
• Let and y = f(u) = sin u
)cossin( xxy
xxu cos
1.cos)sin(cos xxxudx
du
du
dy
dx
dy
)sin)(coscoscos( xxxxx
8
Example- level of
• An environmental study of a certain community suggests that the average daily level of carbon monoxide in the air may be modeled by the formula
2t
175.0)( 2 ppC
parts per million when the population is p thousand. It is estimated that t years from now, the population of the community will be
p(t) = 3.1 + 0.1
2
thousand. At what rate the carbon monoxide level be changing with respect to time 3 years from now.
20C
9
Solution
tppdt
dp
dp
dC
dt
dC2.0)2)(5.0(175.0
2
12
12
When t = 3, p(3) = 3.1 + 0.1(3) 42
24.0)3(2.0)4()174.5.0(2
1 2
12
3
tdt
dC
The carbon monoxide level will be changing at the rate of 0.24 parts per million. It will be increasing because the sign of is positive.
dt
dC
10
Example: Tumor Growth
• A tumor grows such that its radius expands at a constant rate k. Determine the rate of growth of the volume of the tumor when the radius is one centimeter. Assume the shape of the tumor is well approximated by a sphere.
• (Volume of sphere, )
• Solution• Volume of a sphere with radius r is• r and V change with time, and given
• So,
• When r = 1 cm,
3
3
4rV
3
3
4rV
kdt
dr
krkrdt
dr
dr
dV
dt
dV 22 433
4
kkrdt
dV 44 2
11
Example: spider web
• A spider moves horizontally across the ground at a constant rate k, pulling a thin thread with it. One end of the thread is pinned to a vertical wall at height h above the ground and does not move. The other end moves with the spider. Determine the rate of the elongation of the thread.
• Solution
h = height of pinned pointx = position of spiderl = length of thread
h
x
l
12
Spiderweb solution
• From diagram of Pythagoras,
• We know h is constant and rate of spider moves with time is
• Need to find rate of elongation of thread w.r.t. t i.e
• So,
222 hxl k
dt
dx
?dt
dl
dt
dx
l
x
dt
dl
022 dt
dxx
dt
dll
22 hx
xkkl
x
13
Implicit Differentiation
• The equation y = explicitly defines f(x) =
21 x 21 x
as a function of x for 11 x
The same function can also be defined implicitly by the equation
122 yx as long as we restrict y by 10 yTo find the derivative of the explicit form, we use the chain rule;
2
2
122
122
1)2()1(
2
1)1(1
x
xxxx
dx
dx
dx
d
The derivative of the implicit function is
)1()( 22
dx
dyx
dx
d
14
Implicit Differentiation
• So, 022 dx
dyyx
21 x
x
y
x
dx
dy
The procedure illustrated above is called implicit differentiation
Consider another example;
xyyx 332 56
)()(5)6()()( 32332 xdx
dy
dx
d
dx
dx
dx
dyy
dx
dx
1350)2(3 2322
dx
dyyxy
dx
dyyx
15
Continue…
• The derivative involves both x and y and that is acceptable
3222 21)153 xydx
dyyy
222
3
153
21
yyx
xy
dx
dy
16
Example
• Find y' if
Solution
17
Example
• Prove that an equation of the tangent line to the graph of the hyperbola
at the point is
Solution
So the equation of the tangent line at the point P(x0,y0) is
),( 00 yxP
18
Continue…
• So the equation of the tangent line at the point is
Point P is on hyperbola so,
And equation of tangent line is
),( 00 yxP
19
• Find all points (x, y) on the graph of
• (see diagram.) where lines tangent to the graph at (x, y) have slope -1 .
83/23/2 yx
20
Solution
83/23/2 yx
03
2
3
2 3/13/1
dx
dyyx
13/1
3/1
3/1
3/1
x
y
y
x
dx
dy
3/1
3/1
y
x
dx
dy
21
Continue…
• Solve the equation and get the points on the
• curve where the tangent passing through those points have slope -1
• Points: (8,8) and (-8,-8)
13/1
3/1
x
y
dx
dy
22
DERIVATIVE FORMULAS FOR THE INVERSE
TRIGONOMETRIC FUNCTIONS • If u is differentiable function of x, then
dx
du
uu
dx
d2
1
1
1)(sin
dx
du
uu
dx
d2
1
1
1)(cos
dx
du
uu
dx
d2
1
1
1)(tan
dx
du
uuu
dx
d
1||
1)(sec
2
1
dx
du
uu
dx
d2
1
1
1)(cot
dx
du
uuu
dx
d
1||
1)csc
2
1
23
Proof of
• We shall prove the first formula and leave the others as problems.
• Let , u = f(x)
•
dx
du
uu
dx
d2
1
1
1)(sin
)(sin 1 xf)(sin xf
)()(sin xdx
d
dx
d
1cos dx
d
22 )]([1
1
sin1
1
cos
1
xfdx
d
dx
du
udx
duu
du
du
dx
d2
11
1
1)(sin)(sin
)1sin(cos 22
24
Example
• Find the derivative of the following functions:
• 1.
• 2.
• 3.
)12(sin 1 xy
1tan 21 xy
)2(csc 21 xy
25
Solution
• 1. , Let u = (2x + 1)
• 2. ,
• Try 3…
)12(sin 1 xy
xxxdx
duu
du
d
dx
dy
44
2)2(
)12(1
1)(sin
22
1
12 xu
1)1(1
1)2()1(
2
1
)1(1
1)(tan
222
12
21
x
x
xxx
xdx
duu
du
d
dx
dy
1)2( 22
xx
x
1tan 21 xy
26
Differentiating Logarithmic Functions with Bases other than e
• If u = f(x) is a function of x, and is a logarithm with
base b,
Proof: Change to base e and take ln on both sides of and differentiate
Example
Solution
uy blog
dx
du
ubu
dx
db
1.
ln
1)(log
Find the derivative of
u' = 2x
dx
du
du
dy
dx
dy
xx
21
3
7ln
12
12 xu
17ln
62
x
x
ublogb
uy
ln
ln
)1(log3 27 xy
uy ln7ln
3
27
Example
• Find the derivative of 3 ln xy + sin y =
• Solution
2x
This is an implicit function.
28
Another example
• Find the derivative of
• Solution
xxy sin
29
Other Formulas for Derivatives of Exponential Functions
• If u is a function of x, we can obtain the derivative of
an expression in the form :
ue
If we have an exponential function with some base b, we have the following derivative
uby buy lnln
dx
dub
dx
dy
yln
1
30
Example
• Find the derivative of
• Solution
2
10xy
b = 10 and u = 2x
10lnln 2xy
10ln21
xdx
dy
y
10ln2102
xdx
dy x
31
Example
• Show that satisfies the equation
Solution
32
Example
• So
33
Logarithmic differentiation
• Find the derivative of
• Solution
• Complete the steps…• Remember ln e = 1
)74()5(
)12(23
62
xx
xey
x
)74()5(
)12(lnln
23
62
xx
xey
x
)74ln()5ln(2)12ln(6ln2 3 xxxex
)7(
74
1)3(
5
12)2(
12
162
1 23 x
xxxdx
dy
y
34
Related rates and Applications
• Example 1
• A spherical balloon is being filled with a gas in such a way that when the radius is 2ft, the radius is increasing at the rate of 1/6ft./min. How fast is the volume changing at this time?
• (Volume of sphere is )
• Solution Need to find at the
• time when r= 2 and
3
3
4rV
3
3
4rV
3
3
4r
dt
d
dt
dV
dt
drr 24
dt
dV
6
1
dt
dr
6
1)2(4 2
2 rdt
dV
3
8
35
Example 2
A person 6ft. tall walks away from a streetlight at the rate of 5 ft/s.
If the light is 18ft above ground level, how fast is the person’s shadow lengthening
SolutionLet x = length of the shadow y = distance of the person from the streetlight
sftdt
dy/5 ?
dt
dx
Because of similar triangles,
18
6
y x
186
yxx
36
Continue…
186
yxx
xy 2
dt
dx
dt
dy2
dt
dx25
5.2dt
dx The shadow is lengthening at the rate of 2.5ft/s.
37
Example 3
• Consider a piece of ice melting in the shape of a sphere that is melting at a rate of ( Volume of sphere is )
• (a) Model the volume of ice by a function of the radius r
• (b) how fast is the radius changing at the instant when the radius is 4 inch
• (c) how fast is the surface area changing at the same instant
• (surface area of sphere is )
• (d) what assumption are you making in this model about the shape of the ice.
min/5 3in 3
3
4rV
24 rS
38
Solution of melting ice
• Given and r = 4
• (a) Model of ice takes the shape of sphere, so
• (b)
• the radius is decreasing at the rate of
0.025in/min
5dt
dV
3
3
4rV
dt
drr
dt
dV 24
dt
dr2)4(45
025.0~64
5
dt
dr
39
Continue…
• (c)
• With r = 4 and
surface area is decreasing at the rate of
(d) Assumption: Ice is in the shape of a sphere
24 rS
dt
drr
dt
dS 8
025.0~64
5
dt
dr
min/5.264
5)4(8 2in
dt
dS
min/5.2 2in
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