3.6 The chain rule (Page 3/6)

Calculus volume 1 Page 3 / 6

Using the chain rule on a general cosine function

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

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

\begin{matrix} h^{'} (x) & = f^{'} (g (x)) g^{'} (x) & Apply the chain rule. \\ = − sin (g (x)) g^{'} (x) & Substitute f^{'} (g (x)) = − sin (g (x)) . \end{matrix}

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

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

Using the chain rule on a cosine function

Find the derivative of $h (x) = cos (5 x^{2}) .$

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

\begin{matrix} h^{'} (x) & = − sin (5 x^{2}) \cdot 10 x \\ = −10 x sin (5 x^{2}) . \end{matrix}

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Using the chain rule on another trigonometric function

Find the derivative of $h (x) = sec (4 x^{5} + 2 x) .$

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

h^{'} (x) = sec (g (x) tan (g (x)) g^{'} (x) .

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

\begin{matrix} h^{'} (x) & = sec (4 x^{5} + 2 x) tan (4 x^{5} + 2 x) (20 x^{4} + 2) \\ = (20 x^{4} + 2) sec (4 x^{5} + 2 x) tan (4 x^{5} + 2 x) . \end{matrix}

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

$h^{'} (x) = 7 cos (7 x + 2)$

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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 [link] and [link] . 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.

Using the chain rule with trigonometric functions

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

\begin{matrix} \frac{d}{d x} (sin (g (x)) = cos (g (x)) g' (x) & \frac{d}{d x} sin u = cos u \frac{d u}{d x} \\ \frac{d}{d x} (cos (g (x)) = − sin (g (x)) g' (x) & \frac{d}{d x} cos u = − sin u \frac{d u}{d x} \\ \frac{d}{d x} (tan (g (x)) = {sec}^{2} (g (x)) g' (x) & \frac{d}{d x} tan u = {sec}^{2} u \frac{d u}{d x} \\ \frac{d}{d x} (cot (g (x)) = − {csc}^{2} (g (x)) g' (x) & \frac{d}{d x} cot u = − {csc}^{2} u \frac{d u}{d x} \\ \frac{d}{d x} (sec (g (x)) = sec (g (x) tan (g (x)) g' (x) & \frac{d}{d x} sec u = sec u tan u \frac{d u}{d x} \\ \frac{d}{d x} (csc (g (x)) = − csc (g (x)) cot (g (x)) g' (x) & \frac{d}{d x} csc u = − csc u cot u \frac{d u}{d x} . \end{matrix}

Combining the chain rule with the product rule

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

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

\begin{matrix} h^{'} (x) & = \frac{d}{d x} ({(2 x + 1)}^{5}) \cdot {(3 x - 2)}^{7} + \frac{d}{d x} ({(3 x - 2)}^{7}) \cdot {(2 x + 1)}^{5} & Apply the product rule. \\ = 5 {(2 x + 1)}^{4} \cdot 2 \cdot {(3 x - 2)}^{7} + 7 {(3 x - 2)}^{6} \cdot 3 \cdot {(2 x + 1)}^{5} & Apply the chain rule. \\ = 10 {(2 x + 1)}^{4} {(3 x - 2)}^{7} + 21 {(3 x - 2)}^{6} {(2 x + 1)}^{5} & Simplify. \\ = {(2 x + 1)}^{4} {(3 x - 2)}^{6} (10 (3 x - 7) + 21 (2 x + 1)) & Factor out {(2 x + 1)}^{4} {(3 x - 2)}^{6} . \\ = {(2 x + 1)}^{4} {(3 x - 2)}^{6} (72 x - 49) & Simplify. \end{matrix}

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

$h^{'} (x) = \frac{3 - 4 x}{{(2 x + 3)}^{4}}$

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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

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

Then, applying the chain rule once we obtain

k^{'} (x) = \frac{d}{d x} (h (f (g (x))) = h' (f (g (x))) \cdot \frac{d}{d x} f ((g (x))) .

Applying the chain rule again, we obtain

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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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