Exercise 5.2 · Q6: Differentiate $\cos x^3\cdot\sin^2(x^5)$

Basics you need

Solution

1. Read the function and name the two factors

   $$y = \underbrace{\cos(x^3)}_{u}\cdot\underbrace{\big(\sin(x^5)\big)^2}_{v}.$$

   Both factors are defined for every real $x$⁠, and (as the next steps show) each has a derivative at every real $x$⁠. So $\frac{dy}{dx}$ exists for all $x \in \R$⁠.

   The symbol $\cos x^3$ could tempt you to read $(\cos x)^3$⁠. The book's convention is that a power written on the variable belongs to the variable, so $\cos x^3 = \cos(x^3)$⁠; the power written on the function name, as in $\sin^2$⁠, belongs to the function. The textbook answer, which contains $\sin x^3$ and $\cos x^5$⁠, confirms this reading.
   $\sin^2\theta$ is short for $(\sin\theta)^2$⁠, exactly as in $\sin^2\theta + \cos^2\theta = 1$⁠. Basics: trigonometric identities

2. Apply the product rule

   $$\frac{dy}{dx} = u'\,v + u\,v'.$$

   So we need $u'$ and $v'$ separately.

   The last operation in building $y$ is the multiplication of $u$ by $v$⁠, so the outermost rule is the product rule. The chain rule is then used inside it, once for each factor.
   Product (Leibnitz) rule, rule (2), Section 5.3. Basics: derivative rules

3. Differentiate $u = \cos(x^3)$ (two links)

   Inner function $t = x^3$⁠; outer function $\cos t$⁠.

   $$\frac{dt}{dx} = 3x^2, \qquad \frac{du}{dt} = -\sin t.$$ $$\begin{aligned} u' &= \frac{du}{dt}\cdot\frac{dt}{dx} = (-\sin t)(3x^2) \\ &= -3x^2\sin x^3. \end{aligned}$$
   The cosine sees $x^3$⁠, not $x$⁠. So its rate is $-\sin$ evaluated at $x^3$⁠, and then we multiply by how fast $x^3$ itself changes, $3x^2$⁠. Writing $-\sin x^3$ alone would forget the inner rate.
   Chain rule, Theorem 4, Section 5.3.1; $(\cos t)' = -\sin t$ and $(x^3)' = 3x^2$ from Table 5.3. Lesson: the chain rule

4. Differentiate $v = \big(\sin(x^5)\big)^2$ (three links)

   Name the links from the inside out: $s = x^5$⁠, then $w = \sin s$⁠, then $v = w^2$⁠.

   $$\frac{ds}{dx} = 5x^4, \qquad \frac{dw}{ds} = \cos s, \qquad \frac{dv}{dw} = 2w.$$ $$\begin{aligned} v' &= \frac{dv}{dw}\cdot\frac{dw}{ds}\cdot\frac{ds}{dx} \\ &= 2\sin x^5\cdot\cos x^5\cdot 5x^4 \\ &= 10x^4\sin x^5\cos x^5. \end{aligned}$$
   Three operations are stacked: raise $x$ to the fifth power, take the sine, square. Each one contributes one factor: $2w$ from squaring (with $w = \sin x^5$⁠), $\cos s$ from the sine (with $s = x^5$⁠), and $5x^4$ from the fifth power.
   Chain rule extended to three functions (Section 5.3.1, after Theorem 4); $(w^2)' = 2w$⁠, $(\sin s)' = \cos s$⁠, $(x^5)' = 5x^4$ from Table 5.3. Basics: composite functions

5. Put everything into the product rule

   $$\begin{aligned} \frac{dy}{dx} &= u'v + uv' \\ &= (-3x^2\sin x^3)\,\sin^2 x^5 \\ &\quad + \cos x^3\,(10x^4\sin x^5\cos x^5). \end{aligned}$$

   Writing the positive term first:

   $$\begin{aligned} \frac{dy}{dx} &= 10x^4\sin x^5\cos x^5\cos x^3 \\ &\quad - 3x^2\sin x^3\sin^2 x^5. \end{aligned}$$
   Each term of the product rule keeps one factor as it is and replaces the other by its derivative: $u'v$ differentiates the cosine factor, $uv'$ differentiates the $\sin^2$ factor. Reordering the terms changes nothing, but it makes the answer match the textbook's layout.
   Optional tidy form: $2\sin x^5\cos x^5 = \sin 2x^5$⁠, so the first term is also $5x^4\sin(2x^5)\cos x^3$⁠. Basics: trigonometric identities

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