Exercise 5.5 · Q11: Differentiate $(x\cos x)^x+(x\sin x)^{1/x}$

Basics you need

Solution

1. Where are both bases positive?

   We need $x\cos x > 0$ and $x\sin x > 0$ (and $x \ne 0$ for the exponent $\frac1x$⁠).

   • If $x > 0$⁠: we need $\cos x > 0$ and $\sin x > 0$⁠, so $x$ is in a first-quadrant interval $(2k\pi,\ 2k\pi + \frac{\pi}{2})$⁠, $k = 0, 1, 2, \ldots$
   • If $x < 0$⁠: we need $\cos x < 0$ and $\sin x < 0$⁠, so $x$ is in a third-quadrant interval. The negative ones are
   $$\left(-(2k+1)\pi,\ -(2k+1)\pi + \tfrac{\pi}{2}\right),$$

   $k = 0, 1, 2, \ldots$⁠; for $k = 0$ this is $(-\pi, -\frac{\pi}{2})$⁠.

   On these intervals both terms are defined and positive. Put

   $$u = (x\cos x)^x, \qquad v = (x\sin x)^{1/x},$$

   so $y = u + v$ and $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$⁠.

   A product of two numbers is positive exactly when they have the same sign. For $x > 0$ both $\cos x$ and $\sin x$ must be positive; for $x < 0$ both must be negative. For example $x = -2.5$ lies in $(-\pi, -\frac{\pi}{2})$⁠, where $\cos x$ and $\sin x$ are both negative, so $x\cos x = 2.0028590 > 0$ and $x\sin x = 1.4961804 > 0$⁠. The book does not state the domain; the formula below holds on all of these intervals.
   Signs of $\sin$ and $\cos$ by quadrant; positivity requirement (Section 5.5). Basics: trigonometric identities

2. $\frac{du}{dx}$ for $u = (x\cos x)^x$

   Take logs: $\log u = x\log(x\cos x)$⁠. We need the derivative of $\log(x\cos x)$⁠. With $g = x\cos x$ and $g' = \cos x - x\sin x$ (product rule):

   $$\begin{aligned} \frac{d}{dx}\log(x\cos x) &= \frac{\cos x - x\sin x}{x\cos x} \\ &= \frac1x - \tan x. \end{aligned}$$

   Now the product rule on $x\cdot\log(x\cos x)$⁠:

   $$\begin{aligned} \frac1u\frac{du}{dx} &= 1\cdot\log(x\cos x) + x\left(\frac1x - \tan x\right) \\ &= 1 - x\tan x + \log(x\cos x). \end{aligned}$$

   Multiply by $u$⁠:

   $$\begin{aligned} \frac{du}{dx} = \ &(x\cos x)^x \\ &\times\big[1 - x\tan x + \log(x\cos x)\big]. \end{aligned}$$
   We keep $\log(x\cos x)$ as the log of one positive quantity. Writing it as $\log x + \log\cos x$ would be wrong on the negative intervals, where $x$ and $\cos x$ are both negative and neither has a logarithm. Dividing term by term, $\frac{\cos x}{x\cos x} = \frac1x$ and $\frac{x\sin x}{x\cos x} = \tan x$ (both allowed, since $x \ne 0$ and $\cos x \ne 0$ here).
   $\frac{d}{dx}\log g = \frac{g'}{g}$ for $g > 0$ (Theorem 5 and chain rule); product rule (Section 5.3). Lesson: logarithmic differentiation

3. $\frac{dv}{dx}$ for $v = (x\sin x)^{1/x}$

   Take logs: $\log v = \frac1x\log(x\sin x)$⁠. With $g = x\sin x$⁠, $g' = \sin x + x\cos x$⁠:

   $$\begin{aligned} \frac{d}{dx}\log(x\sin x) &= \frac{\sin x + x\cos x}{x\sin x} \\ &= \frac1x + \cot x. \end{aligned}$$

   Product rule on $\frac1x\cdot\log(x\sin x)$⁠, using $\frac{d}{dx}\frac1x = -\frac1{x^2}$⁠:

   $$\begin{aligned} \frac1v\frac{dv}{dx} &= -\frac{1}{x^2}\log(x\sin x) + \frac1x\left(\frac1x + \cot x\right) \\ &= \frac{x\cot x + 1 - \log(x\sin x)}{x^2}. \end{aligned}$$

   Multiply by $v$⁠:

   $$\begin{aligned} \frac{dv}{dx} = \ &(x\sin x)^{\frac1x} \\ &\times\left[\frac{x\cot x + 1 - \log(x\sin x)}{x^2}\right]. \end{aligned}$$
   The exponent $\frac1x$ comes down as a factor. To combine the three terms over the common denominator $x^2$⁠: $\frac1x\cdot\frac1x = \frac{1}{x^2}$ and $\frac1x\cot x = \frac{x\cot x}{x^2}$⁠. Here $\sin x \ne 0$⁠, so $\cot x$ is defined.
   $(x^{-1})' = -x^{-2}$ (Table 5.3); product rule; $\frac{d}{dx}\log g = \frac{g'}{g}$⁠. Basics: derivative rules

4. Add the two derivatives

   $$\begin{aligned} \frac{dy}{dx} = \ &(x\cos x)^x\big[1 - x\tan x + \log(x\cos x)\big] \\ &+ (x\sin x)^{\frac1x}\left[\frac{x\cot x + 1 - \log(x\sin x)}{x^2}\right], \end{aligned}$$

   valid on every interval of Step 1.

   Sum rule. Each part was derived using only $x\cos x > 0$ and $x\sin x > 0$⁠, so the result holds wherever the function itself is defined.
   Sum rule (Section 5.3). Basics: derivative rules

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