Ulearngo

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 \(h \left(\right. x \left.\right) = \text{cos} \left(\right. g \left(\right. x \left.\right) \left.\right) .\)

Solution

Think of \(h \left(\right. x \left.\right) = \text{cos} \left(\right. g \left(\right. x \left.\right) \left.\right)\) as \(f \left(\right. g \left(\right. x \left.\right) \left.\right)\) where \(f \left(\right. x \left.\right) = \text{cos} x .\) Since \(f^{'} \left(\right. x \left.\right) = − \text{sin} x .\) we have \(f^{'} \left(\right. g \left(\right. x \left.\right) \left.\right) = − \text{sin} \left(\right. g \left(\right. x \left.\right) \left.\right) .\) Then we do the following calculation.

\[\begin{aligned} h^{'} \left(\right. x \left.\right) & = f^{'} \left(\right. g \left(\right. x \left.\right) \left.\right) g^{'} \left(\right. x \left.\right) & & & \text{Apply the chain rule}. \\ & = − \text{sin} \left(\right. g \left(\right. x \left.\right) \left.\right) g^{'} \left(\right. x \left.\right) & & & \text{Substitute} f^{'} \left(\right. g \left(\right. x \left.\right) \left.\right) = − \text{sin} \left(\right. g \left(\right. x \left.\right) \left.\right) . \end{aligned}\]

Thus, the derivative of \(h \left(\right. x \left.\right) = \text{cos} \left(\right. g \left(\right. x \left.\right) \left.\right)\) is given by \(h^{'} \left(\right. x \left.\right) = − \text{sin} \left(\right. g \left(\right. x \left.\right) \left.\right) g^{'} \left(\right. x \left.\right) .\)

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 \(h \left(\right. x \left.\right) = \text{cos} \left(\right. 5 x^{2} \left.\right) .\)

Solution

Let \(g \left(\right. x \left.\right) = 5 x^{2} .\) Then \(g^{'} \left(\right. x \left.\right) = 10 x .\) Using the result from the previous example,

\[\begin{aligned} h^{'} \left(\right. x \left.\right) & = − \text{sin} \left(\right. 5 x^{2} \left.\right) \cdot 10 x \\ & = −10 x \text{sin} \left(\right. 5 x^{2} \left.\right) . \end{aligned}\]

Example 3.53

Using the Chain Rule on Another Trigonometric Function

Find the derivative of \(h \left(\right. x \left.\right) = \text{sec} \left(\right. 4 x^{5} + 2 x \left.\right) .\)

Solution

Apply the chain rule to \(h \left(\right. x \left.\right) = \text{sec} \left(\right. g \left(\right. x \left.\right) \left.\right)\) to obtain

\[h^{'} \left(\right. x \left.\right) = \text{sec} \left(\right. g \left(\right. x \left.\right) \left.\right) \text{tan} \left(\right. g \left(\right. x \left.\right) \left.\right) g^{'} \left(\right. x \left.\right) .\]

In this problem, \(g \left(\right. x \left.\right) = 4 x^{5} + 2 x ,\) so we have \(g^{'} \left(\right. x \left.\right) = 20 x^{4} + 2 .\) Therefore, we obtain

\[\begin{aligned} h^{'} \left(\right. x \left.\right) & = \text{sec} \left(\right. 4 x^{5} + 2 x \left.\right) \text{tan} \left(\right. 4 x^{5} + 2 x \left.\right) \left(\right. 20 x^{4} + 2 \left.\right) \\ & = \left(\right. 20 x^{4} + 2 \left.\right) \text{sec} \left(\right. 4 x^{5} + 2 x \left.\right) \text{tan} \left(\right. 4 x^{5} + 2 x \left.\right) . \end{aligned}\]

Checkpoint 3.36

Find the derivative of \(h \left(\right. x \left.\right) = \text{sin} \left(\right. 7 x + 2 \left.\right) .\)

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 \(x\) for which the derivative is defined,

\[\begin{aligned} \frac{d}{d x} \left(\right. \text{sin} \left(\right. g \left(\right. x \left.\right) \left.\right) \left.\right) = \text{cos} \left(\right. g \left(\right. x \left.\right) \left.\right) g ' \left(\right. x \left.\right) & & & \frac{d}{d x} \text{sin} u = \text{cos} u \frac{d u}{d x} \\ \frac{d}{d x} \left(\right. \text{cos} \left(\right. g \left(\right. x \left.\right) \left.\right) \left.\right) = − \text{sin} \left(\right. g \left(\right. x \left.\right) \left.\right) g ' \left(\right. x \left.\right) & & & \frac{d}{d x} \text{cos} u = − \text{sin} u \frac{d u}{d x} \\ \frac{d}{d x} \left(\right. \text{tan} \left(\right. g \left(\right. x \left.\right) \left.\right) \left.\right) = \text{sec}^{2} \left(\right. g \left(\right. x \left.\right) \left.\right) g ' \left(\right. x \left.\right) & & & \frac{d}{d x} \text{tan} u = \text{sec}^{2} u \frac{d u}{d x} \\ \frac{d}{d x} \left(\right. \text{cot} \left(\right. g \left(\right. x \left.\right) \left.\right) \left.\right) = − \text{csc}^{2} \left(\right. g \left(\right. x \left.\right) \left.\right) g ' \left(\right. x \left.\right) & & & \frac{d}{d x} \text{cot} u = − \text{csc}^{2} u \frac{d u}{d x} \\ \frac{d}{d x} \left(\right. \text{sec} \left(\right. g \left(\right. x \left.\right) \left.\right) \left.\right) = \text{sec} \left(\right. g \left(\right. x \left.\right) \text{tan} \left(\right. g \left(\right. x \left.\right) \left.\right) g ' \left(\right. x \left.\right) & & & \frac{d}{d x} \text{sec} u = \text{sec} u \text{tan} u \frac{d u}{d x} \\ \frac{d}{d x} \left(\right. \text{csc} \left(\right. g \left(\right. x \left.\right) \left.\right) \left.\right) = − \text{csc} \left(\right. g \left(\right. x \left.\right) \left.\right) \text{cot} \left(\right. g \left(\right. x \left.\right) \left.\right) g ' \left(\right. x \left.\right) & & & \frac{d}{d x} \text{csc} u = − \text{csc} u \text{cot} u \frac{d u}{d x} . \end{aligned}\]

Example 3.54

Combining the Chain Rule with the Product Rule

Find the derivative of \(h \left(\right. x \left.\right) = \left(\right. 2 x + 1 \left.\right)^{5} \left(\right. 3 x - 2 \left.\right)^{7} .\)

Solution

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

\[\begin{aligned} h^{'} \left(\right. x \left.\right) & = \frac{d}{d x} \left(\right. \left(\left(\right. 2 x + 1 \left.\right)\right)^{5} \left.\right) \cdot \left(\right. 3 x - 2 \left.\right)^{7} + \frac{d}{d x} \left(\right. \left(\left(\right. 3 x - 2 \left.\right)\right)^{7} \left.\right) \cdot \left(\right. 2 x + 1 \left.\right)^{5} & & & \text{Apply the product rule}. \\ & = 5 \left(\left(\right. 2 x + 1 \left.\right)\right)^{4} \cdot 2 \cdot \left(\right. 3 x - 2 \left.\right)^{7} + 7 \left(\left(\right. 3 x - 2 \left.\right)\right)^{6} \cdot 3 \cdot \left(\left(\right. 2 x + 1 \left.\right)\right)^{5} & & & \text{Apply the chain rule}. \\ & = 10 \left(\right. 2 x + 1 \left.\right)^{4} \left(\right. 3 x - 2 \left.\right)^{7} + 21 \left(\left(\right. 3 x - 2 \left.\right)\right)^{6} \left(\left(\right. 2 x + 1 \left.\right)\right)^{5} & & & \text{Simplify}. \\ & = \left(\right. 2 x + 1 \left.\right)^{4} \left(\right. 3 x - 2 \left.\right)^{6} \left(\right. 10 \left(\right. 3 x - 2 \left.\right) + 21 \left(\right. 2 x + 1 \left.\right) \left.\right) & & & \text{Factor out} \left(\right. 2 x + 1 \left.\right)^{4} \left(\right. 3 x - 2 \left.\right)^{6} . \\ & = \left(\right. 2 x + 1 \left.\right)^{4} \left(\right. 3 x - 2 \left.\right)^{6} \left(\right. 72 x + 1 \left.\right) & & & \text{Simplify}. \end{aligned}\]

Checkpoint 3.37

Find the derivative of \(h \left(\right. x \left.\right) = \frac{x}{\left(\right. 2 x + 3 \left.\right)^{3}} .\)

This lesson is part of:

Derivatives

View Full Tutorial

Track Your Learning Progress

Sign in to unlock unlimited practice exams, tutorial practice quizzes, personalized weak area practice, AI study assistance with Lexi, and detailed performance analytics.