270.8TheChainRule

8/13/2019 270.8TheChainRule

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The 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 ishelpful 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 thederivative of the outside leaving the inside unchanged times thederivative 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 thenusing 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 thederivative using the Chain Rule.



Using the Chain Rule, it will be helpful to identify the outside andinside 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 x x x x x x x xh

Now, we would not normally expand that derivative. It was doneonly 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

)(' x g

dx

du ))((')(' x g f u f

du

dy

)('))((' x g x g f dx

dy

dx

du

du

dy

dx

dy

Comparing the two representations:



Another example: Find the derivative of 32 )3( x x f

It is helpful to identify the outside function and the insidefunction. 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 functionleaving the inside function unchanged and then multiply by thederivative of the inside function.

x x x f 233' 22

The derivative of theinside using the PowerRule

The derivative ofthe outside leavingthe inside unchanged

3



32 )3( x x f

Finally we need to simplify the answer.

So the solution to finding the derivative of

is . 22 36' x x x f



Example 3: Find the derivative of 3 2 13 x xh

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

13 2 x

3

12 13 x xh

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

32

13

1

3

1

3

1

3

1

withStarting

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

outside function.



x x xh 61331' 3

22

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

3 22

3

22

13

2or 132'

x

x x x xh

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

The derivative of theoutside leaving theinside unchanged

The derivative of theinside



Example 4: Find the derivative of: 2

14

2)(

x x f

This could be done by the quotient rule, but the numerator is aconstant whose derivative is 0. Since the numerator does notcontain a variable we could just do a rewrite and use the chainrule.

2

2 142

14

2)(

x

x x f

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

41422)(

142)(

3

2

x x f

x x f

Don’t mess with the inside!!!

derivative of the inside



3

3

3

14

16'

or

1416)('

41422)(

x x f

x x f

x x f

The result needs some simplifying:

Multiply the constants together.



Example 5: Find the derivative of:x4

3xxg

)(

The outside is , the inside is .

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

2

1

x4

3xxg

)(

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

of .

x

x

4

3

x

x

4

3



x

x

dx

d

x

x x g

4

3

4

3

2

1)('

2

1

2

1

4

3)(

x

x x g

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

x x

x

x x g

x

x x

x

x

x g

Continued…



3

4

42

1)('

3

4

42

1)('

4

1

4

3

2

1)('

2

21

2

2

2

1

x

x

x x g

x

x

x x g

x x

x x g

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 thebinomial. It is not asking for a derivative.

You can useFOIL to find the

productFirstOutsideInsideLast



2

f (x) 2x 6 9

Suppose you did want to find the derivative of thisproblem.

Your first step would be to rewrite the radical asan exponential

Step 1 2

1

2f (x) 2x 6 9

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

PowerRule

ChainRule

ProductRule

Click on the arrow of the answer you would pick.



2

1

2f (x) 2x 6 9

If you picked the chain rule, you are correct.

2 1

1 1

2 2d

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

Derivative of Outside times Derivative of Inside

(Power Rule)

1

1 11

2 21 d

f (x) 2 2x 6 9 2x 6 2x 6 02 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 1

12 2

1

2

1

2

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

22

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

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

2x 6

Now to simplify using our algebra skills.



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

2x 62x 6 2x 62 2x 6 9 2x 6 2x 6 9 2x 6

f (x)2(x 3) (x 3)

Yeah! Its finished.



You may have noticed in the last example that when youare using several rules for differentiating in combination,the algebra can get very complicated. That is why it isessential that before you begin a problem you havedetermined HOW you are going to find the derivativeand 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 areclose because when you perform the chain

rule on the outside, you will be using thepower 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’treally have a product unless you write it as a

product

Product

To use the productrule you would haveto start with this.

Go back andchoose again.

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