Exercise 5.2 · Q1: Differentiate $\sin(x^2+5)$

Basics you need

Solution

1. Name the inner and the outer function

   Let $y = \sin(x^2+5)$ and put

   $$\begin{aligned} t &= x^2 + 5 &&(\text{inner}), \\ y &= \sin t &&(\text{outer}). \end{aligned}$$

   Substituting the first into the second gives back $y = \sin(x^2+5)$⁠, so this split describes our function exactly.

   Table 5.3 tells us how to differentiate $\sin t$⁠, but only when the input is the variable we are differentiating with respect to. Here the input of sine is $x^2+5$⁠, not $x$⁠. Giving the input its own name $t$ turns one unfamiliar derivative into two familiar ones.
   To find the inner function, ask: "what would I compute first on a calculator?" For $\sin(x^2+5)$ you would compute $x^2+5$ first and press sin last, so $x^2+5$ is inner and sine is outer. Basics: composite functions

2. Differentiate the outer function with respect to $t$

   $$\frac{dy}{dt} = \frac{d}{dt}(\sin t) = \cos t.$$
   Inside this link of the chain, $t$ is the variable and sine is applied directly to it, so the standard result applies with no adjustment. This derivative exists for every real $t$⁠.
   $\frac{d}{dt}(\sin t) = \cos t$ is in Table 5.3 (Section 5.3), with the letter $t$ in place of $x$⁠. The name of the variable does not matter. Basics: derivative rules

3. Differentiate the inner function with respect to $x$

   $$\frac{dt}{dx} = \frac{d}{dx}\left(x^2\right) + \frac{d}{dx}(5) = 2x + 0 = 2x.$$
   The sum rule lets us differentiate $x^2$ and $5$ separately. The power rule with $n=2$ gives $2x^{1} = 2x$⁠. The constant 5 does not change when $x$ changes, so its rate of change is $0$⁠. This derivative also exists for every real $x$⁠.
   Power rule: $\frac{d}{dx}(x^n) = nx^{n-1}$⁠. Sum rule: $(u+v)' = u' + v'$⁠. Constant: $\frac{d}{dx}(c) = 0$⁠. All three are in Section 5.3 (rule (1) and Table 5.3). Basics: derivative rules

4. Multiply the two rates (chain rule) and return to $x$

   Both derivatives exist for every real number, so the chain rule applies everywhere:

   $$\begin{aligned} \frac{dy}{dx} &= \frac{dy}{dt}\cdot\frac{dt}{dx} \\ &= \cos t \cdot 2x \\ &= 2x\cos\left(x^2+5\right). \end{aligned}$$

   In the last line we replaced $t$ by $x^2+5$⁠, because the answer should be expressed in terms of $x$ only.

   The chain rule multiplies the rates along the chain: a small change $\Delta x$ produces a change in $t$ of about $2x\,\Delta x$⁠, and that produces a change in $y$ of about $\cos t \cdot 2x\,\Delta x$⁠. Dividing by $\Delta x$ gives the product. We write $2x$ in front so that nobody misreads the answer as $\cos(2x\cdot\ldots)$⁠.
   Theorem 4 (Chain Rule), Section 5.3.1: if $f = v\circ u$⁠, $t = u(x)$⁠, and both $\frac{dt}{dx}$ and $\frac{dv}{dt}$ exist, then $\frac{df}{dx} = \frac{dv}{dt}\cdot\frac{dt}{dx}$⁠. The textbook's Example 21 ($\sin x^2$⁠) is the same pattern. Lesson: the chain rule · Example 21

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