the chain rule homework

3.6 The Chain Rule

Learning objectives.

• 3.6.1 State the chain rule for the composition of two functions.
• 3.6.2 Apply the chain rule together with the power rule.
• 3.6.3 Apply the chain rule and the product/quotient rules correctly in combination when both are necessary.
• 3.6.4 Recognize the chain rule for a composition of three or more functions.
• 3.6.5 Describe the proof of the chain rule.

We have seen the techniques for differentiating basic functions ( x n , sin x , cos x , etc . ) ( x n , sin x , cos x , etc . ) as well as sums, differences, products, quotients, and constant multiples of these functions. However, these techniques do not allow us to differentiate compositions of functions, such as h ( x ) = sin ( x 3 ) h ( x ) = sin ( x 3 ) or k ( x ) = 3 x 2 + 1 . k ( x ) = 3 x 2 + 1 . In this section, we study the rule for finding the derivative of the composition of two or more functions.

Deriving the Chain Rule

When we have a function that is a composition of two or more functions, we could use all of the techniques we have already learned to differentiate it. However, using all of those techniques to break down a function into simpler parts that we are able to differentiate can get cumbersome. Instead, we use the chain rule , which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

To put this rule into context, let’s take a look at an example: h ( x ) = sin ( x 3 ) . h ( x ) = sin ( x 3 ) . We can think of the derivative of this function with respect to x as the rate of change of sin ( x 3 ) sin ( x 3 ) relative to the change in x . x . Consequently, we want to know how sin ( x 3 ) sin ( x 3 ) changes as x x changes. We can think of this event as a chain reaction: As x x changes, x 3 x 3 changes, which leads to a change in sin ( x 3 ) . sin ( x 3 ) . This chain reaction gives us hints as to what is involved in computing the derivative of sin ( x 3 ) . sin ( x 3 ) . First of all, a change in x x forcing a change in x 3 x 3 suggests that somehow the derivative of x 3 x 3 is involved. In addition, the change in x 3 x 3 forcing a change in sin ( x 3 ) sin ( x 3 ) suggests that the derivative of sin ( u ) sin ( u ) with respect to u , u , where u = x 3 , u = x 3 , is also part of the final derivative.

We can take a more formal look at the derivative of h ( x ) = sin ( x 3 ) h ( x ) = sin ( x 3 ) by setting up the limit that would give us the derivative at a specific value a a in the domain of h ( x ) = sin ( x 3 ) . h ( x ) = sin ( x 3 ) .

This expression does not seem particularly helpful; however, we can modify it by multiplying and dividing by the expression x 3 − a 3 x 3 − a 3 to obtain

From the definition of the derivative, we can see that the second factor is the derivative of x 3 x 3 at x = a . x = a . That is,

However, it might be a little more challenging to recognize that the first term is also a derivative. We can see this by letting u = x 3 u = x 3 and observing that as x → a , u → a 3 : x → a , u → a 3 :

Thus, h ′ ( a ) = cos ( a 3 ) · 3 a 2 . h ′ ( a ) = cos ( a 3 ) · 3 a 2 .

In other words, if h ( x ) = sin ( x 3 ) , h ( x ) = sin ( x 3 ) , then h ′ ( x ) = cos ( x 3 ) · 3 x 2 . h ′ ( x ) = cos ( x 3 ) · 3 x 2 . Thus, if we think of h ( x ) = sin ( x 3 ) h ( x ) = sin ( x 3 ) as the composition ( f ∘ g ) ( x ) = f ( g ( x ) ) ( f ∘ g ) ( x ) = f ( g ( x ) ) where f ( x ) = f ( x ) = sin x x and g ( x ) = x 3 , g ( x ) = x 3 , then the derivative of h ( x ) = sin ( x 3 ) h ( x ) = sin ( x 3 ) is the product of the derivative of g ( x ) = x 3 g ( x ) = x 3 and the derivative of the function f ( x ) = sin x f ( x ) = sin x evaluated at the function g ( x ) = x 3 . g ( x ) = x 3 . At this point, we anticipate that for h ( x ) = sin ( g ( x ) ) , h ( x ) = sin ( g ( x ) ) , it is quite likely that h ′ ( x ) = cos ( g ( x ) ) g ′ ( x ) . h ′ ( x ) = cos ( g ( x ) ) g ′ ( x ) . As we determined above, this is the case for h ( x ) = sin ( x 3 ) . h ( x ) = sin ( x 3 ) .

Now that we have derived a special case of the chain rule, we state the general case and then apply it in a general form to other composite functions. An informal proof is provided at the end of the section.

Rule: The Chain Rule

Let f f and g g be functions. For all x in the domain of g g for which g g is differentiable at x and f f is differentiable at g ( x ) , g ( x ) , the derivative of the composite function

is given by

Alternatively, if y y is a function of u , u , and u u is a function of x , x , then

Watch an animation of the chain rule.

Problem-Solving Strategy

Problem-solving strategy: applying the chain rule.

• To differentiate h ( x ) = f ( g ( x ) ) , h ( x ) = f ( g ( x ) ) , begin by identifying f ( x ) f ( x ) and g ( x ) . g ( x ) .
• Find f ′ ( x ) f ′ ( x ) and evaluate it at g ( x ) g ( x ) to obtain f ′ ( g ( x ) ) . f ′ ( g ( x ) ) .
• Find g ′ ( x ) . g ′ ( x ) .
• Write h ′ ( x ) = f ′ ( g ( x ) ) · g ′ ( x ) . h ′ ( x ) = f ′ ( g ( x ) ) · g ′ ( x ) .

Note : When applying the chain rule to the composition of two or more functions, keep in mind that we work our way from the outside function in. It is also useful to remember that the derivative of the composition of two functions can be thought of as having two parts; the derivative of the composition of three functions has three parts; and so on. Also, remember that we never evaluate a derivative at a derivative.

The Chain and Power Rules Combined

We can now apply the chain rule to composite functions, but note that we often need to use it with other rules. For example, to find derivatives of functions of the form h ( x ) = ( g ( x ) ) n , h ( x ) = ( g ( x ) ) n , we need to use the chain rule combined with the power rule. To do so, we can think of h ( x ) = ( g ( x ) ) n h ( x ) = ( g ( x ) ) n as f ( g ( x ) ) f ( g ( x ) ) where f ( x ) = x n . f ( x ) = x n . Then f ′ ( x ) = n x n − 1 . f ′ ( x ) = n x n − 1 . Thus, f ′ ( g ( x ) ) = n ( g ( x ) ) n − 1 . f ′ ( g ( x ) ) = n ( g ( x ) ) n − 1 . This leads us to the derivative of a power function using the chain rule,

Rule: Power Rule for Composition of Functions

For all values of x for which the derivative is defined, if

Example 3.48

Using the chain and power rules.

Find the derivative of h ( x ) = 1 ( 3 x 2 + 1 ) 2 . h ( x ) = 1 ( 3 x 2 + 1 ) 2 .

First, rewrite h ( x ) = 1 ( 3 x 2 + 1 ) 2 = ( 3 x 2 + 1 ) −2 . h ( x ) = 1 ( 3 x 2 + 1 ) 2 = ( 3 x 2 + 1 ) −2 .

Applying the power rule with g ( x ) = 3 x 2 + 1 , g ( x ) = 3 x 2 + 1 , we have

Rewriting back to the original form gives us

Checkpoint 3.34

Find the derivative of h ( x ) = ( 2 x 3 + 2 x − 1 ) 4 . h ( x ) = ( 2 x 3 + 2 x − 1 ) 4 .

Example 3.49

Using the chain and power rules with a trigonometric function.

Find the derivative of h ( x ) = sin 3 x . h ( x ) = sin 3 x .

First recall that sin 3 x = ( sin x ) 3 , sin 3 x = ( sin x ) 3 , so we can rewrite h ( x ) = sin 3 x h ( x ) = sin 3 x as h ( x ) = ( sin x ) 3 . h ( x ) = ( sin x ) 3 .

Applying the power rule with g ( x ) = sin x , g ( x ) = sin x , we obtain

Example 3.50

Finding the equation of a tangent line.

Find the equation of a line tangent to the graph of h ( x ) = 1 ( 3 x − 5 ) 2 h ( x ) = 1 ( 3 x − 5 ) 2 at x = 2 . x = 2 .

Because we are finding an equation of a line, we need a point. The x -coordinate of the point is 2. To find the y -coordinate, substitute 2 into h ( x ) . h ( x ) . Since h ( 2 ) = 1 ( 3 ( 2 ) − 5 ) 2 = 1 , h ( 2 ) = 1 ( 3 ( 2 ) − 5 ) 2 = 1 , the point is ( 2 , 1 ) . ( 2 , 1 ) .

For the slope, we need h ′ ( 2 ) . h ′ ( 2 ) . To find h ′ ( x ) , h ′ ( x ) , first we rewrite h ( x ) = ( 3 x − 5 ) −2 h ( x ) = ( 3 x − 5 ) −2 and apply the power rule to obtain

By substituting, we have h ′ ( 2 ) = −6 ( 3 ( 2 ) − 5 ) −3 = −6 . h ′ ( 2 ) = −6 ( 3 ( 2 ) − 5 ) −3 = −6 . Therefore, the line has equation y − 1 = −6 ( x − 2 ) . y − 1 = −6 ( x − 2 ) . Rewriting, the equation of the line is y = −6 x + 13 . y = −6 x + 13 .

Checkpoint 3.35

Find the equation of the line tangent to the graph of f ( x ) = ( x 2 − 2 ) 3 f ( x ) = ( x 2 − 2 ) 3 at x = −2 . x = −2 .

Combining the Chain Rule with Other Rules

Now that we can combine the chain rule and the power rule, we examine how to combine the chain rule with the other rules we have learned. In particular, we can use it with the formulas for the derivatives of trigonometric functions or with the product rule.

Example 3.51

Using the chain rule on a general cosine function.

Find the derivative of h ( x ) = cos ( g ( x ) ) . h ( x ) = cos ( g ( x ) ) .

Think of h ( x ) = cos ( g ( x ) ) h ( x ) = cos ( g ( x ) ) as f ( g ( x ) ) f ( g ( x ) ) where f ( x ) = cos x . f ( x ) = cos x . Since f ′ ( x ) = − sin x . f ′ ( x ) = − sin x . we have f ′ ( g ( x ) ) = − sin ( g ( x ) ) . f ′ ( g ( x ) ) = − sin ( g ( x ) ) . Then we do the following calculation.

Thus, the derivative of h ( x ) = cos ( g ( x ) ) h ( x ) = cos ( g ( x ) ) is given by h ′ ( x ) = − sin ( g ( x ) ) g ′ ( x ) . h ′ ( x ) = − sin ( g ( x ) ) g ′ ( x ) .

In the following example we apply the rule that we have just derived.

Example 3.52

Using the chain rule on a cosine function.

Find the derivative of h ( x ) = cos ( 5 x 2 ) . h ( x ) = cos ( 5 x 2 ) .

Let g ( x ) = 5 x 2 . g ( x ) = 5 x 2 . Then g ′ ( x ) = 10 x . g ′ ( x ) = 10 x . Using the result from the previous example,

Example 3.53

Using the chain rule on another trigonometric function.

Find the derivative of h ( x ) = sec ( 4 x 5 + 2 x ) . h ( x ) = sec ( 4 x 5 + 2 x ) .

Apply the chain rule to h ( x ) = sec ( g ( x ) ) h ( x ) = sec ( g ( x ) ) to obtain

In this problem, g ( x ) = 4 x 5 + 2 x , g ( x ) = 4 x 5 + 2 x , so we have g ′ ( x ) = 20 x 4 + 2 . g ′ ( x ) = 20 x 4 + 2 . Therefore, we obtain

Checkpoint 3.36

Find the derivative of h ( x ) = sin ( 7 x + 2 ) . h ( x ) = sin ( 7 x + 2 ) .

At this point we provide a list of derivative formulas that may be obtained by applying the chain rule in conjunction with the formulas for derivatives of trigonometric functions. Their derivations are similar to those used in Example 3.51 and Example 3.53 . For convenience, formulas are also given in Leibniz’s notation, which some students find easier to remember. (We discuss the chain rule using Leibniz’s notation at the end of this section.) It is not absolutely necessary to memorize these as separate formulas as they are all applications of the chain rule to previously learned formulas.

Theorem 3.10

Using the chain rule with trigonometric functions.

For all values of x x for which the derivative is defined,

Example 3.54

Combining the chain rule with the product rule.

Find the derivative of h ( x ) = ( 2 x + 1 ) 5 ( 3 x − 2 ) 7 . h ( x ) = ( 2 x + 1 ) 5 ( 3 x − 2 ) 7 .

First apply the product rule, then apply the chain rule to each term of the product.

Checkpoint 3.37

Find the derivative of h ( x ) = x ( 2 x + 3 ) 3 . h ( x ) = x ( 2 x + 3 ) 3 .

Composites of Three or More Functions

We can now combine the chain rule with other rules for differentiating functions, but when we are differentiating the composition of three or more functions, we need to apply the chain rule more than once. If we look at this situation in general terms, we can generate a formula, but we do not need to remember it, as we can simply apply the chain rule multiple times.

In general terms, first we let

Then, applying the chain rule once we obtain

Applying the chain rule again, we obtain

Rule: Chain Rule for a Composition of Three Functions

For all values of x for which the function is differentiable, if

In other words, we are applying the chain rule twice.

Notice that the derivative of the composition of three functions has three parts. (Similarly, the derivative of the composition of four functions has four parts, and so on.) Also, remember, we can always work from the outside in, taking one derivative at a time.

Example 3.55

Differentiating a composite of three functions.

Find the derivative of k ( x ) = cos 4 ( 7 x 2 + 1 ) . k ( x ) = cos 4 ( 7 x 2 + 1 ) .

First, rewrite k ( x ) k ( x ) as

Then apply the power rule several times.

Checkpoint 3.38

Find the derivative of h ( x ) = sin 6 ( x 3 ) . h ( x ) = sin 6 ( x 3 ) .

Example 3.56

Using the chain rule in a velocity problem.

A particle moves along a coordinate axis. Its position at time t is given by s ( t ) = sin ( 2 t ) + cos ( 3 t ) . s ( t ) = sin ( 2 t ) + cos ( 3 t ) . What is the velocity of the particle at time t = π 6 ? t = π 6 ?

To find v ( t ) , v ( t ) , the velocity of the particle at time t , t , we must differentiate s ( t ) . s ( t ) . Thus,

Substituting t = π 6 t = π 6 into v ( t ) , v ( t ) , we obtain v ( π 6 ) = −2 . v ( π 6 ) = −2 .

Checkpoint 3.39

A particle moves along a coordinate axis. Its position at time t t is given by s ( t ) = sin ( 4 t ) . s ( t ) = sin ( 4 t ) . Find its acceleration at time t . t .

At this point, we present a very informal proof of the chain rule. For simplicity’s sake we ignore certain issues: For example, we assume that g ( x ) ≠ g ( a ) g ( x ) ≠ g ( a ) for x ≠ a x ≠ a in some open interval containing a . a . We begin by applying the limit definition of the derivative to the function h ( x ) h ( x ) to obtain h ′ ( a ) : h ′ ( a ) :

Rewriting, we obtain

Although it is clear that

it is not obvious that

To see that this is true, first recall that since g is differentiable at a , g a , g is also continuous at a . a . Thus,

Next, make the substitution y = g ( x ) y = g ( x ) and b = g ( a ) b = g ( a ) and use change of variables in the limit to obtain

Example 3.57

Using the chain rule with functional values.

Let h ( x ) = f ( g ( x ) ) . h ( x ) = f ( g ( x ) ) . If g ( 1 ) = 4 , g ′ ( 1 ) = 3 , g ( 1 ) = 4 , g ′ ( 1 ) = 3 , and f ′ ( 4 ) = 7 , f ′ ( 4 ) = 7 , find h ′ ( 1 ) . h ′ ( 1 ) .

Use the chain rule, then substitute.

Checkpoint 3.40

Given h ( x ) = f ( g ( x ) ) . h ( x ) = f ( g ( x ) ) . If g ( 2 ) = −3 , g ′ ( 2 ) = 4 , g ( 2 ) = −3 , g ′ ( 2 ) = 4 , and f ′ ( −3 ) = 7 , f ′ ( −3 ) = 7 , find h ′ ( 2 ) . h ′ ( 2 ) .

The Chain Rule Using Leibniz’s Notation

As with other derivatives that we have seen, we can express the chain rule using Leibniz’s notation. This notation for the chain rule is used heavily in physics applications.

For h ( x ) = f ( g ( x ) ) , For h ( x ) = f ( g ( x ) ) , let u = g ( x ) u = g ( x ) and y = h ( x ) = f ( u ) . y = h ( x ) = f ( u ) . Thus,

Consequently,

Rule: Chain Rule Using Leibniz’s Notation

If y y is a function of u , u , and u u is a function of x , x , then

Example 3.58

Taking a derivative using leibniz’s notation, example 1.

Find the derivative of y = ( x 3 x + 2 ) 5 . y = ( x 3 x + 2 ) 5 .

First, let u = x 3 x + 2 . u = x 3 x + 2 . Thus, y = u 5 . y = u 5 . Next, find d u d x d u d x and d y d u . d y d u . Using the quotient rule,

Finally, we put it all together.

It is important to remember that, when using the Leibniz form of the chain rule, the final answer must be expressed entirely in terms of the original variable given in the problem.

Example 3.59

Taking a derivative using leibniz’s notation, example 2.

Find the derivative of y = tan ( 4 x 2 − 3 x + 1 ) . y = tan ( 4 x 2 − 3 x + 1 ) .

First, let u = 4 x 2 − 3 x + 1 . u = 4 x 2 − 3 x + 1 . Then y = tan u . y = tan u . Next, find d u d x d u d x and d y d u : d y d u :

Checkpoint 3.41

Use Leibniz’s notation to find the derivative of y = cos ( x 3 ) . y = cos ( x 3 ) . Make sure that the final answer is expressed entirely in terms of the variable x . x .

Section 3.6 Exercises

For the following exercises, given y = f ( u ) y = f ( u ) and u = g ( x ) , u = g ( x ) , find d y d x d y d x by using Leibniz’s notation for the chain rule: d y d x = d y d u d u d x . d y d x = d y d u d u d x .

y = 3 u − 6 , u = 2 x 2 y = 3 u − 6 , u = 2 x 2

y = 6 u 3 , u = 7 x − 4 y = 6 u 3 , u = 7 x − 4

y = sin u , u = 5 x − 1 y = sin u , u = 5 x − 1

y = cos u , u = − x 8 y = cos u , u = − x 8

y = tan u , u = 9 x + 2 y = tan u , u = 9 x + 2

y = 4 u + 3 , u = x 2 − 6 x y = 4 u + 3 , u = x 2 − 6 x

For each of the following exercises,

• decompose each function in the form y = f ( u ) y = f ( u ) and u = g ( x ) , u = g ( x ) , and
• find d y d x d y d x as a function of x . x .

y = ( 3 x − 2 ) 6 y = ( 3 x − 2 ) 6

y = ( 3 x 2 + 1 ) 3 y = ( 3 x 2 + 1 ) 3

y = sin 5 ( x ) y = sin 5 ( x )

y = ( x 7 + 7 x ) 7 y = ( x 7 + 7 x ) 7

y = tan ( sec x ) y = tan ( sec x )

y = csc ( π x + 1 ) y = csc ( π x + 1 )

y = cot 2 x y = cot 2 x

y = −6 ( sin   x ) - 3 y = −6 ( sin   x ) - 3

For the following exercises, find d y d x d y d x for each function.

y = ( 3 x 2 + 3 x − 1 ) 4 y = ( 3 x 2 + 3 x − 1 ) 4

y = ( 5 − 2 x ) −2 y = ( 5 − 2 x ) −2

y = cos 3 ( π x ) y = cos 3 ( π x )

y = ( 2 x 3 − x 2 + 6 x + 1 ) 3 y = ( 2 x 3 − x 2 + 6 x + 1 ) 3

y = 1 sin 2 ( x ) y = 1 sin 2 ( x )

y = ( tan x + sin x ) −3 y = ( tan x + sin x ) −3

y = x 2 cos 4 x y = x 2 cos 4 x

y = sin ( cos 7 x ) y = sin ( cos 7 x )

y = 6 + sec π x 2 y = 6 + sec π x 2

y = cot 3 ( 4 x + 1 ) y = cot 3 ( 4 x + 1 )

Let y = [ f ( x ) ] 2 y = [ f ( x ) ] 2 and suppose that f ′ ( 1 ) = 4 f ′ ( 1 ) = 4 and d y d x = 10 d y d x = 10 for x = 1 . x = 1 . Find f ( 1 ) . f ( 1 ) .

Let y = ( f ( x ) + 5 x 2 ) 4 y = ( f ( x ) + 5 x 2 ) 4 and suppose that f ( −1 ) = −4 f ( −1 ) = −4 and d y d x = 3 d y d x = 3 when x = −1 . x = −1 . Find f ′ ( −1 ) f ′ ( −1 )

Let y = ( f ( u ) + 3 x ) 2 y = ( f ( u ) + 3 x ) 2 and u = x 3 − 2 x . u = x 3 − 2 x . If f ( 4 ) = 6 f ( 4 ) = 6 and d y d x = 18 d y d x = 18 when x = 2 , x = 2 , find f ′ ( 4 ) . f ′ ( 4 ) .

[T] Find the equation of the tangent line to y = − sin ( x 2 ) y = − sin ( x 2 ) at the origin. Use a calculator to graph the function and the tangent line together.

[T] Find the equation of the tangent line to y = ( 3 x + 1 x ) 2 y = ( 3 x + 1 x ) 2 at the point ( 1 , 16 ) . ( 1 , 16 ) . Use a calculator to graph the function and the tangent line together.

Find the x x -coordinates at which the tangent line to y = ( x − 6 x ) 8 y = ( x − 6 x ) 8 is horizontal.

[T] Find an equation of the line that is normal to g ( θ ) = sin 2 ( π θ ) g ( θ ) = sin 2 ( π θ ) at the point ( 1 4 , 1 2 ) . ( 1 4 , 1 2 ) . Use a calculator to graph the function and the normal line together.

For the following exercises, use the information in the following table to find h ′ ( a ) h ′ ( a ) at the given value for a . a .

h ( x ) = f ( g ( x ) ) ; a = 0 h ( x ) = f ( g ( x ) ) ; a = 0

h ( x ) = g ( f ( x ) ) ; a = 0 h ( x ) = g ( f ( x ) ) ; a = 0

h ( x ) = ( x 4 + g ( x ) ) −2 ; a = 1 h ( x ) = ( x 4 + g ( x ) ) −2 ; a = 1

h ( x ) = ( f ( x ) g ( x ) ) 2 ; a = 3 h ( x ) = ( f ( x ) g ( x ) ) 2 ; a = 3

h ( x ) = f ( x + f ( x ) ) ; a = 1 h ( x ) = f ( x + f ( x ) ) ; a = 1

h ( x ) = ( 1 + g ( x ) ) 3 ; a = 2 h ( x ) = ( 1 + g ( x ) ) 3 ; a = 2

h ( x ) = g ( 2 + f ( x 2 ) ) ; a = 1 h ( x ) = g ( 2 + f ( x 2 ) ) ; a = 1

h ( x ) = f ( g ( sin x ) ) ; a = 0 h ( x ) = f ( g ( sin x ) ) ; a = 0

[T] The position function of a freight train is given by s ( t ) = 100 ( t + 1 ) −2 , s ( t ) = 100 ( t + 1 ) −2 , with s s in meters and t t in seconds. At time t = 6 t = 6 s, find the train’s

• velocity and
• acceleration.
• Using a. and b. is the train speeding up or slowing down?

[T] A mass hanging from a vertical spring is in simple harmonic motion as given by the following position function, where t t is measured in seconds and s s is in inches:

s ( t ) = −3 cos ( π t + π 4 ) . s ( t ) = −3 cos ( π t + π 4 ) .

• Determine the position of the spring at t = 1.5 t = 1.5 s.
• Find the velocity of the spring at t = 1.5 t = 1.5 s.

[T] The total cost to produce x x boxes of Thin Mint Girl Scout cookies is C C dollars, where C = 0.0001 x 3 − 0.02 x 2 + 3 x + 300 . C = 0.0001 x 3 − 0.02 x 2 + 3 x + 300 . In t t weeks production is estimated to be x = 1600 + 100 t x = 1600 + 100 t boxes.

• Find the marginal cost C ′ ( x ) . C ′ ( x ) .
• Use Leibniz’s notation for the chain rule, d C d t = d C d x · d x d t , d C d t = d C d x · d x d t , to find the rate with respect to time t t that the cost is changing.
• Use b. to determine how fast costs are increasing when t = 2 t = 2 weeks. Include units with the answer.

[T] The formula for the area of a circle is A = π r 2 , A = π r 2 , where r r is the radius of the circle. Suppose a circle is expanding, meaning that both the area A A and the radius r r (in inches) are expanding.

• Suppose r = 2 − 100 ( t + 7 ) 2 r = 2 − 100 ( t + 7 ) 2 where t t is time in seconds. Use the chain rule d A d t = d A d r · d r d t d A d t = d A d r · d r d t to find the rate at which the area is expanding.
• Use a. to find the rate at which the area is expanding at t = 4 t = 4 s.

[T] The formula for the volume of a sphere is S = 4 3 π r 3 , S = 4 3 π r 3 , where r r (in feet) is the radius of the sphere. Suppose a spherical snowball is melting in the sun.

• Suppose r = 1 ( t + 1 ) 2 − 1 12 r = 1 ( t + 1 ) 2 − 1 12 where t t is time in minutes. Use the chain rule d S d t = d S d r · d r d t d S d t = d S d r · d r d t to find the rate at which the snowball is melting.
• Use a. to find the rate at which the volume is changing at t = 1 t = 1 min.

[T] The daily temperature in degrees Fahrenheit of Phoenix in the summer can be modeled by the function T ( x ) = 94 − 10 cos [ π 12 ( x − 2 ) ] , T ( x ) = 94 − 10 cos [ π 12 ( x − 2 ) ] , where x x is hours after midnight. Find the rate at which the temperature is changing at 4 p.m.

[T] The depth (in feet) of water at a dock changes with the rise and fall of tides. The depth is modeled by the function D ( t ) = 5 sin ( π 6 t − 7 π 6 ) + 8 , D ( t ) = 5 sin ( π 6 t − 7 π 6 ) + 8 , where t t is the number of hours after midnight. Find the rate at which the depth is changing at 6 a.m.

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29 Chain Rule

With one additional rule, we will have the power to take the derivative of any function we can write down. What is this amazing rule? Why, it’s called the chain rule . The chain rule is [latex]\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)[/latex]. If we drop the [latex](x)[/latex] on each function (note that it is still there, it is just implied), we have a slightly shorter version:

[latex]\boxed{\cfrac{d}{dx} f(g) = f'(g) \cdot g'}[/latex]

Here, [latex]f(g)[/latex] is called function composition. It does not mean [latex]f[/latex] times [latex]g[/latex]. It means we are sticking [latex]g[/latex] inside of [latex]f[/latex]! Like [latex]f[/latex] is eating [latex]g[/latex]! That’s actually cannibalism if you think about it — so don’t think about it too closely. But do remember how to do function composition.

Why does this work? I’m not going to do a formal proof, but let’s run through an idea behind it. Recall that the derivative is the slope or how steep the graph of a function is. This is a lot easier to think about if we’re talking about lines. For example, suppose [latex]{\color{red}f = 6x - 10}[/latex], and [latex]{\color{blue}g = \frac{1}{2}x + 3}[/latex]. As we know from algebra, the slope of the [latex]{\color{red}f}[/latex] line is [latex]{\color{red}6}[/latex], and the slope of the [latex]{\color{blue}g}[/latex] line is [latex]{\color{blue} \frac{1}{2}}[/latex].

So what is the slope of [latex]{\color{red}f}({\color{blue}g})[/latex]? What this means is we are putting [latex]{\color{blue}g}[/latex] inside of [latex]{\color{red}f}[/latex]. So if [latex]{\color{red}f = 6x - 10}[/latex], and [latex]{\color{blue}g = \frac{1}{2}x + 3}[/latex], we take the blue [latex]{\color{blue}\frac{1}{2}x + 3}[/latex] and use that to replace the red [latex]{\color{red}x}[/latex]. Here is what is would look like: \begin{align*} {\color{red}f}({\color{blue}g}) & = {\color{red}6{\color{blue}\left(\frac{1}{2}x + 3\right)}-10} \\ & = 6\left(\frac{1}{2}x + 3\right)-10 \\ & = 3x + 18 – 10 \\ & = 3x – 8 \end{align*} So again, what is the slope of [latex]{\color{red}f}({\color{blue}g})[/latex]? We can see from this calculation that it is [latex]3[/latex], the product of the two slopes [latex]{\color{red}6}[/latex] and [latex]{\color{blue} \frac{1}{2}}[/latex]. That’s why you have [latex]f' \cdot g'[/latex] in the formula.

Okay, but what about the [latex](g)[/latex] after the [latex]f'[/latex] in the formula? One way to think about the function composition [latex]f(g)[/latex] is we are looking at the [latex]f[/latex] curve at the [latex]x[/latex]-location of [latex]g[/latex]. When you take the derivative, you’re still looking at that same location (now on the [latex]f'[/latex] curve), so you still need that [latex](g)[/latex] there to specify that location.

We must identify an “inside” (g) and “outside” (f) function in order to use the chain rule. Often, the “inside” function will be in parentheses (though not always). That works in this case, so the “inside” function is [latex]x^2 + x[/latex], so [latex]g = x^2 + x[/latex]. The outside function is [latex]\ln[/latex], and hence [latex]f =\ln(x)[/latex]. We also know [latex]f' = \frac{1}{x}[/latex] and [latex]g' = 2x + 1[/latex]. Now to use the chain rule, we first need [latex]f'(g)[/latex]. What is this? Well, remember that this is not multiplication, but it is sticking one function inside another. In this case, we are taking [latex]g = x^2 + x[/latex] and sticking it into the function [latex]f' = \frac{1}{x}[/latex]. This means we replace the [latex]x[/latex] in [latex]\frac{1}{x}[/latex], replacing it with [latex]x^2 + x[/latex], and get [latex]f'(g) = \frac{1}{x^2 + x}[/latex]. Hence we have \begin{align*} \frac{d}{dx} \ln(x^2 + x) & = f'(g) \cdot g’ \\ & = \frac{1}{(x^2 + x)} \cdot (2x + 1) \\ & = \boxed{\frac{2x + 1}{x^2 + x}}. \end{align*} There you have it!

Using the power rule, we first multiply [latex](3x + 1)^2 = (3x + 1)(3x + 1) = 9x^2 + 3x + 3x + 1 = 9x^2 + 6x + 1[/latex]. In this form, it is easy to find the derivative: [latex]\frac{d}{dx} 9x^2 + 6x + 1 = 18x + 6[/latex].

Using the chain rule, we identify the inside [latex]g[/latex] function as [latex]3x + 1[/latex], and the outside function as [latex]x^2[/latex]. We then have \begin{align*} \frac{d}{dx} (3x + 1)^2 & = f'(g) \cdot g’ \\ & = 2(3x + 1)(3) \\ & = 6(3x + 1) \\ & = \boxed{18x + 6}. \end{align*} Again, math just works!

Things can get quite complicated with the chain rule.

There are no explicit parentheses here, but the square root acts like parentheses, and it designates an inside function of [latex]g = x^2 + 2e^x[/latex]. The outside function is therefore [latex]f = \sqrt[4]{x}[/latex]. If we rewrite [latex]\sqrt[4]{x}[/latex] as [latex]x^{1/4}[/latex], we can use the power rule and find [latex]f' = \frac{1}{4} x^{-3/4}[/latex]. We also have [latex]g' = 2x + 2e^x[/latex]. Hence, the chain rule gives \begin{align*} \frac{d}{dx} \sqrt[4]{ x^2 + 2e^x} & = f'(g) \cdot g’ \\ & = \frac{1}{4} (x^2 + 2e^x)^{-3/4} (2x + 2e^x) \\ & = \frac{2x + 2e^x}{4} \frac{1}{\sqrt[4]{x^2 + 2e^x}^3} \\ & = \boxed{\frac{x + e^x}{2 \sqrt[4]{x^2 + 2e^x}^3 }}. \end{align*} That’s as simplified as we can get the answer to be.

One more quick example.

Recall that [latex]e^{\ln(a)} = a[/latex]. To prove this rule, we rewrite [latex]a^x = (e^{\ln(a)})^x = e^{x \ln(a)}[/latex]. We are then computing

[latex]\frac{d}{dx} \ a^x = \frac{d}{dx} \ e^{x \ln(a)}[/latex]

To compute this derivative, we set [latex]f = e^x[/latex] and [latex]g = x \ln(a)[/latex]. We find [latex]f' = e^x[/latex] and [latex]g' = \ln(a)[/latex], so by the chain rule \begin{align*} \frac{d}{dx} \ a^x & = \frac{d}{dx} \ e^{x \ln(a)} \\ & = f'(g) \cdot g’ \\ & = e^{x \ln(a)} \cdot \ln(a) \\ \end{align*} We’ve already shown that [latex]a^x = e^{x \ln(a)}[/latex], so this simplifies to [latex]a^x \cdot \ln(a)[/latex], as desired.

Informal Calculus Copyright © by Tyler Seacrest is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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3.12: Homework- Chain Rule

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• Page ID 88652

• Tyler Seacrest
• University of Montana Western
• Watch this video from Khan Academy: Chain Rule Definition and example
• \(a(x) = (x^2 + 5)^{20}\) \(40 x (x^2 + 5)^{19}\) ans
• \(b(x) = e^{x^2}\) \(2x e^{x^2}\) ans
• \(c(x) = (kx + r)^n\) for constants \(k\), \(r\), \(n\). \(k n (kx + r)^{n-1}\) ans
• \(d(x) = (\ln(x))^3 + \ln(x^3)\) \(\frac{3 (\ln(x))^2}{x} + \frac{3}{x}\) ans
• \(e(x) = \sin(\cos(x))\) \(-\sin(x) \cos(\cos(x))\) ans
• \(f(x) = e^{\sin(x) + \cos(x)}\) \((\cos(x) - \sin(x)) e^{\sin(x) + \cos(x)}\) ans
• \(g(x) = \sqrt{3x^2 - 5x + 6}\) \(\frac{6x - 5}{2 \sqrt{3x^2 - 5x + 6}}\) ans
• \(h(x) = e^{-x}\) \(-e^{-x}\) ans
• \(\frac{d}{dx} \ln(x^3)\) \(\frac{3}{x}\) ans
• \(\frac{d}{dx} \ln(x e^x)\) \(\frac{1}{x} + 1\) ans
• Use logarithm rules to explain why \(\frac{d}{dx} \ln(e^5 \cdot x) = \frac{d}{dx} \ln(x)\). Using logarithm rules, we have that \(ln(e^5 \cdot x) = \ln(x) + \ln(e^5) = \ln(x) + 5\). This has the same derivative as \(\ln(x)\) since we are just adding a constant. ans
• Simplify \(e^{\ln(2)}\) \(=2\) ans
• Simplify \(\ln(e^2)\). \(=2\) ans
• Simplify \(e^{\ln(2) + x}\) \(2e^x\) ans
• Simplify \((e^{\ln(2) x})\) \(2^x\) ans
• Use part (d) to compute \(\frac{d}{dx} 2^x\). \(\ln(2) 2^x\) ans