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

We have seen the techniques for differentiating basic functions ( 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 ) or 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 ) . We can think of the derivative of this function with respect to x as the rate of change of sin ( x 3 ) relative to the change in x . Consequently, we want to know how sin ( x 3 ) changes as x changes. We can think of this event as a chain reaction: As x changes, x 3 changes, which leads to a change in sin ( x 3 ) . This chain reaction gives us hints as to what is involved in computing the derivative of sin ( x 3 ) . First of all, a change in x forcing a change in x 3 suggests that somehow the derivative of x 3 is involved. In addition, the change in x 3 forcing a change in sin ( x 3 ) suggests that the derivative of sin ( u ) with respect to u , where 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 ) by setting up the limit that would give us the derivative at a specific value a in the domain of h ( x ) = sin ( x 3 ) .

h ( a ) = lim x a sin ( x 3 ) sin ( a 3 ) x a .

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

h ( a ) = lim x a sin ( x 3 ) sin ( a 3 ) x 3 a 3 · x 3 a 3 x a .

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

lim x a x 3 a 3 x a = d d x ( x 3 ) = 3 a 2 .

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 and observing that as x a , u a 3 :

lim x a sin ( x 3 ) sin ( a 3 ) x 3 a 3 = lim u a 3 sin u sin ( a 3 ) u a 3 = d d u ( sin u ) u = a 3 = cos ( a 3 ).

Thus, h ( a ) = cos ( a 3 ) · 3 a 2 .

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

Questions & Answers

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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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