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

h ( x ) = ( f g ) ( x ) = f ( g ( x ) )

is given by

h ( x ) = f ( g ( x ) ) g ( x ) .

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

d y d x = d y d u · d u d x .

Watch an animation of the chain rule.

Problem-solving strategy: applying the chain rule

  1. To differentiate h ( x ) = f ( g ( x ) ) , begin by identifying f ( x ) and g ( x ) .
  2. Find f ( x ) and evaluate it at g ( x ) to obtain f ( g ( x ) ) .
  3. Find g ( x ) .
  4. Write 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 , we need to use the chain rule combined with the power rule. To do so, we can think of h ( x ) = ( g ( x ) ) n as f ( g ( x ) ) where f ( x ) = x n . Then f ( x ) = n x n 1 . Thus, f ( g ( x ) ) = n ( g ( x ) ) n 1 . This leads us to the derivative of a power function using the chain rule,

h ( x ) = n ( g ( x ) ) n 1 g ( x )

Rule: power rule for composition of functions

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

h ( x ) = ( g ( x ) ) n .

Then

h ( x ) = n ( g ( x ) ) n 1 g ( x ).

Using the chain and power rules

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

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

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

h ( x ) = −2 ( 3 x 2 + 1 ) −3 ( 6 x ) .

Rewriting back to the original form gives us

h ( x ) = −12 x ( 3 x 2 + 1 ) 3 .
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Find the derivative of h ( x ) = ( 2 x 3 + 2 x 1 ) 4 .

h ( x ) = 4 ( 2 x 3 + 2 x 1 ) 3 ( 6 x + 2 ) = 8 ( 3 x + 1 ) ( 2 x 3 + 2 x 1 ) 3

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Using the chain and power rules with a trigonometric function

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

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

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

h ( x ) = 3 ( sin x ) 2 cos x = 3 sin 2 x cos x .
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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 at 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 ) . Since h ( 2 ) = 1 ( 3 ( 2 ) 5 ) 2 = 1 , the point is ( 2 , 1 ) .

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

h ( x ) = −2 ( 3 x 5 ) −3 ( 3 ) = −6 ( 3 x 5 ) −3 .

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

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

y = −48 x 88

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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.

Practice Key Terms 1

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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