2.4 the chain rule greg kelly, hanford high school, richland, washingtonphoto by vickie kelly, 2002

2.4 The Chain Rule

Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2002

Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2002

U.S.S. AlabamaMobile, Alabama

HWQ

Let f(x) and g(x) be 2 differentiable functions such that:

x F(x) G(x) F’(x) G’(x)

4 1 7 8 -8

3 -5 -3 -4 6

-5 2 -10 9 -1

Find the derivative of f(x)g(x) at x = -5. -92

Calculus Warm-Up

3cosd

xdx

23cos sinx x

Calculus Warm-Up

2 1d

xdx

We will come back to this problem later.

The Chain Rule

Copyright © Cengage Learning. All rights reserved.

2.4

Find the derivative of a composite function using the Chain Rule.

Objective:

We now have a pretty good list of “shortcuts” to find derivatives of simple functions.

Of course, many of the functions that we will encounter are not so simple. What is needed is a way to combine derivative rules to evaluate more complicated functions.

We do this with the chain rule.

Consider a simple composite function:

6 10y x

2 3 5y x

If 3 5u x

then 2y u

6 10y x 2y u 3 5u x

6dy

dx 2

dy

du 3

du

dx

dy dy du

dx du dx

6 2 3

and another:

5 2y u

where 3u t

then 5 3 2y t

3u t

15dy

dt 5

dy

du 3

du

dt

dy dy du

dt du dt

15 5 3

5 3 2y t

15 2y t

5 2y u

and another:29 6 1y x x

23 1y x

If 3 1u x

3 1u x

18 6dy

xdx

2dy

udu

3du

dx

dy dy du

dx du dx

2y u

2then y u

29 6 1y x x

2 3 1dy

xdu

6 2dy

xdu

18 6 6 2 3x x This pattern is called the chain rule.

dy dy du

dx du dx Chain Rule:

Example: sinf x x 2 4g x x Find: at 2f g x

( ( )) '( ( )) '( )df g x f g x g x

dx

2sin 4f g x x

or:

2sin 4y f g x x

2 2cos 4 4dy d

x xdx dx

2cos 4 2dy

x xdx

Differentiate the outside function...

…then the inside function

at 2, 4x y

( ( )) '( ( )) '( )df g x f g x g x

dxChain Rule:

2cos 2 4 2 2dy

dx

cos 0 4dy

dx

4dy

dx

22 cos 4dy

x xdx

Use the chain rule to differentiate:

3cosd

xdx

23cos sinx x

Use the chain rule to differentiate:

2 1d

xdx

2 1

x

x

Another example:

2cos 3d

xdx

2cos 3

dx

dx

2 cos 3 cos 3d

x xdx

derivative of theoutside function

derivative of theinside function

It looks like we need to use the chain rule again!

con’t:

2cos 3d

xdx

2cos 3

dx

dx

2 cos 3 cos 3d

x xdx

2cos 3 sin 3 3d

x x xdx

2cos 3 sin 3 3x x

6cos 3 sin 3x x

The chain rule can be used more than once.

(That’s what makes the “chain” in the “chain rule”!)

The most common mistake on derivative tests is to forget to use the chain rule.

Every derivative problem could be thought of as a chain-rule problem:

2dx

dx2d

x xdx

2 1x 2x

derivative of outside function

derivative of inside function

The derivative of x is one.

Practice:

Differentiate: 32 2f x x

226 2f x x x

Practice:

Differentiate: 2

7

2 3g t

t

3

28

2 3g t

t

sin 2 ' ?

' cos 2 2

' 2cos 2y x

y x y

y x

2

2

tan 3 ' ?

' sec 3 3

' 3sec 3

y x y

y x

y x

' sin 1

cos 1 ' ?

' sin 1 1

y x y

y x

y x

2cos 3 ' ?

' 2cos 3

' 2 cos 3

y x y

y x

y x

2

2

2

cos 3 ' ?

' sin 3

' 6 sin 3

6

y x y

y x x

y x x

2

2

2

cos 3 ' ?

' sin 3 2 3

' 18 si

3

n 9

y x y

x

x x

x

y

y

BC Homework

• 2.4 Day 1 p. 137: 7-31 odd, 41-57 odd,

67-71 odd, 81,83

• 2.4 Day 2: MMM pgs. 44-46

• 2.4 Day 3: MMM pg. 50

2.4 The Chain Rule – Day 2

Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2002

HWQ

Differentiate:

3sin 4f t t

212sin 4 cos 4f t t t

2.4 Warm-up

223

1/

2

322 / 32

3 2

1 Where does ' 0?

Where does ' not exist?

2' 1 2

34

' 0@ 0

' when 1

1

3 1

0 1

f x x f x

f x

f x x x

x

x

f

f x x

x x

f x DNE x x

223 1

f x x

2 2

1/ 2 1/ 22 2 2

1/ 2 1/ 22 3 2

1/ 2 1/ 22 2 3

1/ 2 1/ 22 2

2 3

2

2

3

1/ 2

2

2

1 ' ?

1' 2 1 1 2

2

2

2 3

1 1

2 1 1

1 1

2 1

1

1

1

f x x x f x

f x x x x x x

x x x x

x x x x

x x

x x xx

f x x

x

x

x

x

Common Denominator

3 2

1/ 3 2 / 32 2

2 / 32

21/ 32

2 / 32

2 / 32

1/ 3 2 / 32 2 2

2 / 3 2 / 32 2

2 / 32

2 2

2

1/ 32

2

4

/ 32 2 2

2 / 3 4 /32 32/2

' ?4

11 4 4 2

3'4

24

3 4

4

3 4 4 2

3 4 3 4

4

3 4 2

3 4 3 4 12

3

2

4 3 4 4

4

xf x f x

x

x x x xf x

x

xx

x

x

x x x

x x

x

x x

x x x

x x

x

x

xf

x

x

HWQ (no calculator)

Determine the point(s) at which the graph of

has a horizontal tangent. 2 1

xf x

x

1,1

2

2

2

2

2 22

2 2

2 22 32

3 1 ' ?

3

3 3 3 1 23 1' 2

3 3

3 1 3 2 3 19 6 22

3

3 2

33

9

xy y

x

x x xxy

x x

x x x x x

x

x x

xx

AB Homework

• 2.4 Day 1 p. 137: 1-31 odd, 41-57 odd

• 2.4 Day 2: p. 137: 59-73 odd, 79-89 odd

• 2.4 Day 3: MMM pgs. 44-46

• 2.4 Day 4: Chain Rule W/S