The Chain Rule: Learn It 3

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The Chain Rule: Learn It 3

Applying the Chain Rule Multiple Times

We can integrate the chain rule with other differentiation rules to handle functions composed of multiple functions. When differentiating compositions that involve three or more functions, it’s often necessary to apply the chain rule multiple times. This allows us to systematically derive the derivative without memorizing complex formulas, as the application of the chain rule can be repeated as needed.

In a general case, consider a function $k(x)=h(f(g(x)))$ .

First, apply the chain rule to find:

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

$k^{\prime}(x)=\frac{d}{dx}(h(f(g(x))))=h^{\prime}(f(g(x))) \cdot \frac{d}{dx}(f(g(x)))$

Applying the chain rule once again gives us:

k^{'} (x) = h^{'} (f (g (x))) f^{'} (g (x)) g^{'} (x)

$k^{\prime}(x)=h^{\prime}(f(g(x)))f^{\prime}(g(x))g^{\prime}(x)$

chain tule for a composition of three functions

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

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

$k(x)=h(f(g(x)))$ ,

then

k^{'} (x) = h^{'} (f (g (x))) f^{'} (g (x)) g^{'} (x)

$k^{\prime}(x)=h^{\prime}(f(g(x)))f^{\prime}(g(x))g^{\prime}(x)$

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 always work from the outside in, taking one derivative at a time.

Find the derivative of $k(x)=\cos^4 (7x^2+1)$

First, rewrite $k(x)$ as

k (x) = (\cos (7 x^{2} + 1))^{4}

$k(x)=(\cos(7x^2+1))^4$

Then apply the chain rule several times.

\begin{array}{lllll} k^{'} (x) & = 4 (\cos (7 x^{2} + 1))^{3} (\frac{d}{d x} \cos (7 x^{2} + 1)) & Apply the chain rule. \\ = 4 (\cos (7 x^{2} + 1))^{3} (- \sin (7 x^{2} + 1)) (\frac{d}{d x} (7 x^{2} + 1)) & Apply the chain rule. \\ = 4 (\cos (7 x^{2} + 1))^{3} (- \sin (7 x^{2} + 1)) (14 x) & Apply the chain rule. \\ = - 56 x \sin (7 x^{2} + 1) \cos^{3} (7 x^{2} + 1) & Simplify. \end{array}

$\begin{array}{lllll}k^{\prime}(x) & =4(\cos(7x^2+1))^3(\frac{d}{dx}\cos(7x^2+1)) & & & \text{Apply the chain rule.} \\ & =4(\cos(7x^2+1))^3(−\sin(7x^2+1))(\frac{d}{dx}(7x^2+1)) & & & \text{Apply the chain rule.} \\ & =4(\cos(7x^2+1))^3(−\sin(7x^2+1))(14x) & & & \text{Apply the chain rule.} \\ & =-56x \sin(7x^2+1) \cos^3 (7x^2+1) & & & \text{Simplify.} \end{array}$

Don’t forget that ${\cos }^{n}(t)$ is a commonly used shorthand notation for ${\left(\cos \left(t\right)\right)}^{n}$ . When we write it without the shorthand notation, we can clearly see why the chain rule is necessary in these situations.

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

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

v (t) = s^{'} (t) = 2 \cos (2 t) - 3 \sin (3 t)

$v(t)=s^{\prime}(t)=2 \cos(2t)-3 \sin(3t)$ .

Substituting $t=\frac{\pi}{6}$ into $v(t)$ , we obtain $v(\frac{\pi}{6})=-2$ .

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “3.6 The Chain Rule” here (opens in new window).

At this point, we present a very informal proof of the chain rule.

Proof

For simplicity’s sake we ignore certain issues: For example, we assume that $g(x)\ne g(a)$ for $x\ne a$ in some open interval containing $a$ . We begin by applying the limit definition of the derivative to the function $h(x)$ to obtain $h^{\prime}(a)$ :

h^{'} (a) = lim_{x \to a} \frac{f (g (x)) - f (g (a))}{x - a}

$h^{\prime}(a)=\underset{x\to a}{\lim}\dfrac{f(g(x))-f(g(a))}{x-a}$

Rewriting, we obtain

h^{'} (a) = lim_{x \to a} \frac{f (g (x)) - f (g (a))}{g (x) - g (a)} \cdot \frac{g (x) - g (a)}{x - a}

$h^{\prime}(a)=\underset{x\to a}{\lim}\dfrac{f(g(x))-f(g(a))}{g(x)-g(a)} \cdot \dfrac{g(x)-g(a)}{x-a}$

Although it is clear that

lim_{x \to a} \frac{g (x) - g (a)}{x - a} = g^{'} (a)

$\underset{x\to a}{\lim}\dfrac{g(x)-g(a)}{x-a}=g^{\prime}(a)$ ,

it is not obvious that

lim_{x \to a} \frac{f (g (x)) - f (g (a))}{g (x) - g (a)} = f^{'} (g (a))

$\underset{x\to a}{\lim}\dfrac{f(g(x))-f(g(a))}{g(x)-g(a)}=f^{\prime}(g(a))$

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

lim_{x \to a} g (x) = g (a)

$\underset{x\to a}{\lim}g(x)=g(a)$ .

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

lim_{x \to a} \frac{f (g (x)) - f (g (a))}{g (x) - g (a)} = lim_{y \to b} \frac{f (y) - f (b)}{y - b} = f^{'} (b) = f^{'} (g (a))

$\underset{x\to a}{\lim}\dfrac{f(g(x))-f(g(a))}{g(x)-g(a)}=\underset{y\to b}{\lim}\dfrac{f(y)-f(b)}{y-b}=f^{\prime}(b)=f^{\prime}(g(a))$ .

Finally,

h^{'} (a) = lim_{x \to a} \frac{f (g (x)) - f (g (a))}{g (x) - g (a)} \cdot \frac{g (x) - g (a)}{x - a} = f^{'} (g (a)) g^{'} (a)

$h^{\prime}(a)=\underset{x\to a}{\lim}\dfrac{f(g(x))-f(g(a))}{g(x)-g(a)} \cdot \dfrac{g(x)-g(a)}{x-a}=f^{\prime}(g(a))g^{\prime}(a)$

$_\blacksquare$

Let $h(x)=f(g(x))$ . If $g(1)=4, \, g^{\prime}(1)=3$ , and $f^{\prime}(4)=7$ , find $h^{\prime}(1)$ .

Use the chain rule, then substitute.

\begin{array}{lllll} h^{'} (1) & = f^{'} (g (1)) g^{'} (1) & Apply the chain rule. \\ = f^{'} (4) \cdot 3 & Substitute g (1) = 4 and g^{'} (1) = 3. \\ = 7 \cdot 3 & Substitute f^{'} (4) = 7. \\ = 21 & Simplify. \end{array}

$\begin{array}{lllll}h^{\prime}(1) & =f^{\prime}(g(1))g^{\prime}(1) & & & \text{Apply the chain rule.} \\ & =f^{\prime}(4) \cdot 3 & & & \text{Substitute} \, g(1)=4 \, \text{and} \, g^{\prime}(1)=3. \\ & =7 \cdot 3 & & & \text{Substitute} \, f^{\prime}(4)=7. \\ & =21 & & & \text{Simplify.} \end{array}$

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “3.6 The Chain Rule” here (opens in new window).