The Chain Rule By Dr. Julia Arnold. Rule 7: The Chain Rule If h(x) = g(f(x)), then h’(x) = g’(f(x))f’(x). The Chain Rule deals with the idea of composite.

The Chain RuleThe Chain RuleBy

Dr. Julia Arnold

Rule 7: The Chain Rule

If h(x) = g(f(x)), then h’(x) = g’(f(x))f’(x).

The Chain Rule deals with the idea of composite functions and it is helpful to think about an outside and an inside function when using The Chain Rule.

In other words: The derivative when using the Chain Rule is the derivative of the outside leaving the inside unchanged times the derivative of the inside.

Let’s consider the function h(x) = f(g(x)) where f(x) = x4 and g(x) = x + 2, then h(x)= f(g(x))= (x+2)4.

We could find the derivative by expanding (x + 2)4 and then using the Power Rule.

h(x)=f(g(x)) = x4 +8x3 + 24x2 + 32x +16

So then, h’(x) =[f(g(x))]’ = 4x3 + 24x2 +48x +32

This is a perfectly acceptable way to find the derivative, but multiplying out the binomial can be time consuming.So, now that we know the derivative, let’s find the derivative using the Chain Rule.

Using the Chain Rule, it will be helpful to identify the outside and inside before beginning.

h(x) = (x+2)4.

The outside = ( )4 and the inside = x+2. Can you identify these?

Now using the chain rule:

Derivative = derivative of outside leaving inside * the derivative of the inside.

h’(x) = (4(x+2)3)*(1)

h’(x)= 4(x+2)3

Now you may be saying to yourself “Wait those two derivatives

were not the same.” When we found the derivative without the

Chain Rule we came up with h’(x) =[f(g(x))]’ = 4x3 + 24x2 +48x +32.

When we found the derivative using the Chain Rule we came up with

h’(x)= 4(x+2)3. To see that these two are the same you simply need

to expand the second one.

32482448126424' 23233 xxxxxxxxh

Now, we would not normally expand that derivative. It was done only for the purpose of verifying that the derivatives were the same.

Another representation of the Chain Rule is:

If u = g(x) and f(u) = y then

where and

dx

du

du

dy

dx

dy

)(' xgdx

du ))((')(' xgfuf

du

dy

)('))((' xgxgfdx

dy

dx

du

du

dy

dx

dy

Comparing the two representations:

Another example: Find the derivative of 32 )3( xxf

It is helpful to identify the outside function and the inside function. In this example, the outside function is the cube,and the inside function is x2 +3.

The chain rule says take the derivative of the outside function leaving the inside function unchanged and then multiply by the derivative of the inside function.

xxxf 233'22

The derivative of the inside using the PowerRule

The derivative of the outside leaving the inside unchanged

3

32 )3( xxf

Finally we need to simplify the answer.

So the solution to finding the derivative of

is . 22 36' xxxf

Example 3: Find the derivative of 3 2 13 xxh

The outside function is the cube root function and the inside function is . First rewrite the function with rational exponent:

13 2 x

31

2 13 xxh

To find the derivative of the outside, do the Power Rule:

32

13

1

3

1

3

13

1

withStarting

Note: This is just HOW we are finding the derivative of the outside function.

xxxh 6133

1' 3

22

Now do a little simplification: Multiply the 1/3 and the 6x.

3 22

3

22

13

2or132'

x

xxxxh

Now let’s look at the actual derivative using the Chain Rule.

The derivative of the outside leaving the inside unchanged

The derivative of the inside

Example 4: Find the derivative of: 214

2)(

xxf

This could be done by the quotient rule, but the numerator is a constant whose derivative is 0. Since the numerator does not contain a variable we could just do a rewrite and use the chain rule.

2

2 14214

2)(

x

xxf

Since 2 is a constant the derivative of this expression will be determined by the chain rule.

41422)(

142)(3

2

xxf

xxf

Don’t mess with the inside!!!

derivative of the inside

3

3

3

14

16'

or

1416)('

41422)(

xxf

xxf

xxf

The result needs some simplifying:

Multiply the constants together.

Example 5: Find the derivative of: x43x

xg

)(

The outside is , the inside is .

First rewrite the square root as the exponent 1/2.

21

x43x

xg

)(

To find the derivative we will need to use the Chain Rule and when we multiply by the derivative of the inside we will need to use the Quotient Rule to find the derivative

of .

x

x

4

3

x

x

4

3

x

x

dx

d

x

xxg

4

3

4

3

2

1)('

2

1

2

1

4

3)(

x

xxg

Now we need to do the Quotient Rule on the inside.

2

2

1

2

2

1

4

34

4

3

2

1)('

4

)1(314

4

3

2

1)('

x

xx

x

xxg

x

xx

x

xxg

Continued…

3

4

42

1)('

3

4

42

1)('

4

1

4

3

2

1)('

2

2

1

2

2

2

1

x

x

xxg

x

x

xxg

xx

xxg

Move last term to front and combine with 1/2,also use rule of negative exponent to make positive exponent.

Use radical sign for 1/2 exponent.

Simplifying more …

2

2

2x 6 9 2x 6 9 2x 6 9

2x 6 9 2x 6 9 2x 6 81

First Outside Inside Last

2x 6 18 2x 6 81

2x 18 2x 6 87

In this problem you are being asked to multiply or expand the binomial. It is not asking for a derivative.

You can use FOIL to find the product FirstOutsideInsideLast

2f (x) 2x 6 9

Suppose you did want to find the derivative of this problem.

Your first step would be to rewrite the radical as an exponential

Step 1 21

2f (x) 2x 6 9

Step 2 would be to decide what rule applies to this problem.

Power Rule

Chain Rule

Product Rule

Click on the arrow of the answer you would pick.

21

2f (x) 2x 6 9

If you picked the chain rule, you are correct.

2 11 1

2 2d

f (x) 2 2x 6 9 2x 6 9dx

Derivative of Outside times Derivative of Inside

(Power Rule)

11 1

12 2

1 df (x) 2 2x 6 9 2x 6 2x 6 0

2 dx

Chain Rule Derivative of

a const an t

Continuing

11 1

12 2

1 df (x) 2 2x 6 9 2x 6 2x 6 0

2 dx

11 11

2 2

1

2

1

2

1f (x) 2 2x 6 9 2x 6 2

2

2f (x) 2 2x 6 9 2x 6

2

2 2x 6 9f (x) 2 2x 6 9 2x 6

2x 6

Now to simplify using our algebra skills.Al

gebr

a St

uff

2 2x 6 9 2 2x 6 9 2x 62x 6f (x)

2x 62x 6 2x 6

2 2x 6 9 2x 6 2x 6 9 2x 6f (x)

2(x 3) (x 3)

Yeah! Its finished.

Ratio

naliz

ing

the

deno

min

ator

and

Redu

cing

.

You may have noticed in the last example that when you are using several rules for differentiating in combination, the algebra can get very complicated. That is why it is essential that before you begin a problem you have determined HOW you are going to find the derivative and that you KNOW the rules you are using.

2

2 1

f (x) 2x 6 9

Begin 2 2x 6 9 (some other stuff )

Sorry, if you picked the power rule you are close because when you perform the chain rule on the outside, you will be using the power rule.

Go back and pick again.

2

f (x) 2x 6 9 2x 6 9 2x 6 9

Sorry, if you picked the product rule, we don’t really have a product unless you write it as a product

Product

To use the product rule you would have to start with this.

Go back and choose again.