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