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3.6 The Chain Rule Photo by Vickie Kelly, 2002 Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn HS, Dover, Massachusetts

3.6 The Chain Rule Photo by Vickie Kelly, 2002 Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn

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Page 1: 3.6 The Chain Rule Photo by Vickie Kelly, 2002 Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn

3.6 The Chain Rule

Photo by Vickie Kelly, 2002Created by Greg Kelly, Hanford High School, Richland, WashingtonRevised by Terry Luskin, Dover-Sherborn HS, Dover, Massachusetts

Page 2: 3.6 The Chain Rule Photo by Vickie Kelly, 2002 Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn

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

U.S.S. AlabamaMobile, Alabama

Page 3: 3.6 The Chain Rule Photo by Vickie Kelly, 2002 Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn

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.

Page 4: 3.6 The Chain Rule Photo by Vickie Kelly, 2002 Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn

Consider a simple composite function:

6 10y x

3 52y x

nd ,5a 3u x 2( )y u

6 10y x 2( )y u 3 5u x

6dy

dx 2

dy

du 3

du

dx

dy

d

u

u

dy d

dx dx

6 32

( '( ) ( ) *')d

f g x g x xfx

gd

Page 5: 3.6 The Chain Rule Photo by Vickie Kelly, 2002 Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn

and another:

5 2y u

if 3u t

5 3 2( )y t

3u t

15dy

dt 5

dy

du 3

du

dt

15

5 3 2y t

15 2y t

5 2y u

dy

d

u

u

dy d

dx dx

( '( ) ( ) *')d

f g x g x xfx

gd

5 3

Page 6: 3.6 The Chain Rule Photo by Vickie Kelly, 2002 Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn

and one more:29 6 1y x x

23 1y x

3 1u x

3 1u x

18 6dy

xdx

2dy

udu

3du

dx

dy dy du

dx du dx

2y u

2when y u

29 6 1y x x

2 3 1dy

xdu

This pattern is called the chain rule.

dy/dx = 2(3x + 1)1 • 3

or ( ( )) ' ( ) * 'd

f g x f g x g xdx

Page 7: 3.6 The Chain Rule Photo by Vickie Kelly, 2002 Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn

dy dy du

dx du dx Chain Rule:

If is the composite of and , then:

f g y f u u g x

at at g x xf g f g

If f(g(x)) is the composite of y = f(u) and u = g(x), then:

d/dx(f(g(x)) = d/dx f (at g(x)) • d/dx g(at x)

Page 8: 3.6 The Chain Rule Photo by Vickie Kelly, 2002 Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn

dy dy du

dx du dx Chain Rule:

If is the composite of and , then:

f g y f u u g x

at at ( ) ( )( )xg x g xf g f g f g x

example:

sinf x x

Find:

cosf x x

2g x x 2 22g 0 2f g f g

1 4 4f g

' cos 10 (0)f

22 2 4 0g 2 4g x x

( )

2

g xf g f g

a x

x

t

( 22)gf g f g

Page 9: 3.6 The Chain Rule Photo by Vickie Kelly, 2002 Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn

We could also find the derivative at x = 2 this way:

2sin 4f g x x

2sin 4y x

siny u 2 4u x

cosdy

udu

2du

xdx

dy dy du

dx du dx

cos 2dy

u xdx

2cos 4 2dy

x xdx

2cos 2 4 2 2dy

dx

cos 0 4dy

dx

4dy

dx

Page 10: 3.6 The Chain Rule Photo by Vickie Kelly, 2002 Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn

Here is a way to find the derivative by seeing “layers:”

2sin 4y x

2 2' cos 4 4d

y x xdx

2' cos 4 2y x x

Differentiate the outside function,(keep the inner function unchanged...)

…then multiply by the derivative of the inner function

Evaluate this general derivative at 2, to find ' 4x y

Page 11: 3.6 The Chain Rule Photo by Vickie Kelly, 2002 Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn

Another example:

2cos 3d

xdx

2cos 3

dx

dx

2 cos 3 cos 3d

x xdx

derivative of theoutside power function

derivative of theinside trig function

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

Page 12: 3.6 The Chain Rule Photo by Vickie Kelly, 2002 Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn

Another example:

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”!)

Page 13: 3.6 The Chain Rule Photo by Vickie Kelly, 2002 Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn

Each derivative formula will now include the chain rule!

1n nd duu nu

dx dx sin cos

d duu u

dx dx

cos sind du

u udx dx

2tan secd du

u udx dx

et cetera…

Page 14: 3.6 The Chain Rule Photo by Vickie Kelly, 2002 Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn

The most common mistake in differentiating is to forget to use the chain rule.

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

2dx

dx2

dx x

dx 2 1x 2x

derivative of outside function

derivative of inside function

The derivative of x is one.

Page 15: 3.6 The Chain Rule Photo by Vickie Kelly, 2002 Created by Greg Kelly, Hanford High School, Richland, Washington Revised by Terry Luskin, Dover-Sherborn

Don’t forget to use the chain rule!