Chain Rule – Differentiating Composite Functions.

Chain Rule – Differentiating Composite Functions

Chain Rule – Differentiating Composite Functions

The Chain Rule states:

If f(x) and g(x) are differentiable functions, then

d

dxf g x f g x g x

To determine the derivative of f(g(x)), determine

f x , replace x with g(x), then multiply by g x

Chain Rule – Differentiating Composite Functions

Power of a Function

We can verify the chain rule by looking at the Power of a Function rule.

y x 4 52 6

Differentiate:

g x x( ) 4 52This is the composition y=f(g(x)), where

and f x x( ) 6

Use the Chain Rule. d

dxf g x f g x g x

d

dxf g x f g x g x

f x x( ) 6 5

f g x g x x( ) 6 6 4 55 2 5

g x x( ) 8

So, d

dxf g x x x 6 4 5 82 5

Therefore, dy

dxx x 4 8 4 52 5

y x 4 52 6

Power of a Function Continued

Chain Rule – Differentiating Composite FunctionsReciprocal of a Function

d

dxf g x f g x g x

f g x

g x x( )

1 1

3 52 2 2

g x x( ) 6 So, d

dxf g x

xx

1

3 56

2 2

Therefore, dy

dx

x

x

6

3 52 2

You Try: Differentiate yx

1

3 52

This is the composition y=f(g(x)), where

f xx

( ) 1

and

g x x( ) 3 52

Use the Chain Rule.

f xx

( )12

Chain Rule – Differentiating Composite Functions

Square Root Function

d

dxf g x f g x g x

f g xg x x

( )( )

1

2

1

2 1 3

g x x( ) 3 2So, d

dxf g x

xx

1

2 13

3

2

Therefore,

You Try: Differentiate y x 1 3

This is the composition y=f(g(x)), where

f x x( ) and

g x x( ) 1 3

Use the Chain Rule.

f xx

( )1

2

dy

dx

x

x

3

2 1

2

3

Gears

To help understand this on a more practical level, we will look at an example involving gears.

Example: Gears

12

3Which gear does gear #3 rely on directly?

Papa, Mama and Baby Gears

Every time the papa gear goes around once, the mama gear goes around 3 times.

Every time the mama gear goes around once, the baby gear will go around twice.

Every time the papa gear goes around once, the baby gear will go around …

3x2 = 6 times

PapaMama

Baby

So the rate of change of the baby gear with respect to the Papa gear, is the rate of change of the baby gear with respect to the Mama gear times the rate of change of the Mama gear with respect to the Papa gear.

x, g(x), f(g(x))

g x x( ) 3 Let x be the number of times that the blue gear goes around.x

Let g(x) be the number of times the red gear goes around. It depends on how many times the blue gear goes around. So g(x)=3x.

Let f((g(x)) be the number of times the yellow gear goes around. f depends directly on g(x).

f g x( ( ))

Rates of change of the gears.

g x x( ) 3

x

Since g(x)=3x, the rate of change of g(x) with respect to x is 3. That is:

The rate of change of the yellow gear with respect to the red gear is 2. That is:

d

dxg x( ) 3 g x( ) 3or

d

dg xf g x

( )( ( )) 2 or f g x( ( )) 2

Derivative of f with respect to x

g x x( ) 3

x

The rate of change of the yellow gear with respect to the blue gear is given by:

d

dxf g x( ( ))

This is calculated by multiplying the rate of change of the yellow gear with respect to the red gear (2) times the rate of change of the red gear with respect to the blue gear (3). The result is 2x3 =6.

The rate of change of the yellow gear with respect to the blue gear is given by:

d

dxf g x f g x g x( ( )) ( ( )) ( )

d

dxf g x f g x g x( ( )) ( ( )) ( ) 2 3 6

Which we saw was:

f g x g x( ( )) ( ) 2

Other notation

ux

We can simplify the notation by letting x be the number of times the blue gear goes around, letting u be the number of times the red gear goes around an y being the number of times the yellow gear goes around.

dy

dx

dy

du

du

dx

y

The rate of change of the yellow gear with respect to the blue gear is given by: