Exercise 5.5 · Q8: Differentiate $(\sin x)^x+\sin^{-1}\sqrt x$

Basics you need

Solution

1. Domain: $0 < x \le 1$

   • $\sin^{-1}\sqrt x$⁠: $\sqrt x$ needs $x \ge 0$⁠, and $\sin^{-1}$ needs $\sqrt x \le 1$⁠, i.e. $x \le 1$⁠. So $0 \le x \le 1$⁠.
   • $(\sin x)^x$⁠: the exponent varies, so the base must be positive, $\sin x > 0$⁠. On $[0, 1]$ this holds for $0 < x \le 1$⁠, because $0 < 1 < \pi$ and $\sin x > 0$ on $(0, \pi)$⁠. At $x = 0$ the term would be $0^0$⁠, which is undefined.

   So $y$ is defined for $0 < x \le 1$⁠, and there $u = (\sin x)^x > 0$⁠.

   The key gives the derivative with no domain; the square-root term $\sqrt{x - x^2}$ in it already hints at $0 < x < 1$⁠. Finding the domain first tells us which points the formula can possibly be about. ($\sin 1 \approx 0.841$⁠, and $1$ radian is about $57^\circ$⁠, well inside $(0, \pi)$⁠.)
   Domain of $\sin^{-1}$ is $[-1, 1]$ (Chapter 2); $\sin x > 0$ for $0 < x < \pi$⁠. Basics: trigonometric identities

2. $u = (\sin x)^x$ by logarithmic differentiation

   For $0 < x \le 1$⁠, $\log u = x\log\sin x$⁠. Product rule, with $\frac{d}{dx}\log\sin x = \frac{\cos x}{\sin x} = \cot x$ (chain rule, inner $\sin x > 0$⁠):

   $$\begin{aligned} \frac1u\cdot\frac{du}{dx} &= 1\cdot\log\sin x + x\cot x. \end{aligned}$$

   So

   $$\frac{du}{dx} = (\sin x)^x\,(x\cot x + \log\sin x).$$
   The exponent $x$ comes down as a factor; then the product rule handles $x\cdot\log\sin x$⁠. Multiplying by $u$ (positive) isolates $\frac{du}{dx}$⁠. This part is fine at $x = 1$ as well (as a one-sided derivative), since $\sin 1 > 0$⁠.
   $\log p^q = q\log p$⁠; product rule; $\frac{d}{dx}\log g = \frac{g'}{g}$ (Theorems 4 and 5). Lesson: logarithmic differentiation

3. $w = \sin^{-1}\sqrt x$ by the chain rule

   Outer function $\sin^{-1}t$⁠, inner $t = \sqrt x$⁠. For $0 < x < 1$ we have $0 < t < 1$⁠, so both derivatives exist:

   $$\begin{aligned} \frac{dw}{dx} &= \frac{1}{\sqrt{1 - t^2}}\cdot\frac{dt}{dx} \\ &= \frac{1}{\sqrt{1 - x}}\cdot\frac{1}{2\sqrt x} \\ &= \frac{1}{2\sqrt{x - x^2}}. \end{aligned}$$
   $t^2 = (\sqrt x)^2 = x$⁠, so $\sqrt{1 - t^2} = \sqrt{1 - x}$⁠. Then $\sqrt{1 - x}\,\sqrt x = \sqrt{x(1 - x)} = \sqrt{x - x^2}$⁠, allowed because both numbers under the roots are positive. The formula for $\frac{d}{dt}\sin^{-1}t$ needs $-1 < t < 1$⁠; that is why $x = 1$ (where $t = 1$⁠) and $x = 0$ (where $\sqrt x$ has no derivative) are left out.
   $\frac{d}{dt}\sin^{-1}t = \frac{1}{\sqrt{1-t^2}}$ (Section 5.3.3); $\sqrt a\sqrt b = \sqrt{ab}$ for $a, b \ge 0$⁠. Lesson: inverse trig derivatives

4. What happens at the endpoint $x = 1$⁠?

   At $x = 1$ only a left-hand derivative could exist. For $w$⁠, with small $h > 0$⁠, put $\theta = \sin^{-1}\sqrt h$⁠, so $\sin\theta = \sqrt h$ and $\cos\theta = \sqrt{1 - h}$ ($\theta$ is in $\left[0, \frac{\pi}{2}\right]$⁠). Then $\sin^{-1}\sqrt{1-h} = \frac{\pi}{2} - \theta$⁠, and

   $$\frac{w(1) - w(1-h)}{h} = \frac{\theta}{h} \ge \frac{\sqrt h}{h} = \frac{1}{\sqrt h},$$

   which grows without bound as $h \to 0^+$⁠, using $\theta \ge \sin\theta = \sqrt h$⁠.

   So $w$⁠, and hence $y$ (since $u$ has a finite one-sided derivative there), has no derivative at $x = 1$⁠: the graph arrives at $(1, 2.412)$ with a vertical tangent. The formula therefore holds on the open interval $0 < x < 1$⁠. The figure below shows this.
   $\sin^{-1}c = \frac{\pi}{2} - \theta$ when $c = \cos\theta$ with $\theta \in \left[0, \frac{\pi}{2}\right]$ (Chapter 2 principal range); $\sin\theta \le \theta$ for $\theta \ge 0$ (Class XI, the inequality behind $\lim\frac{\sin x}{x} = 1$⁠). Basics: slope at a point

5. Add the two derivatives

   For $0 < x < 1$⁠,

   $$\begin{aligned} \frac{dy}{dx} = \ &(\sin x)^x\,(x\cot x + \log\sin x) \\ &+ \frac{1}{2\sqrt{x - x^2}}. \end{aligned}$$
   The sum rule adds $u'$ (Step 2) and $w'$ (Step 3). The textbook writes the second term as $\frac12\cdot\frac{1}{\sqrt{x - x^2}}$⁠, which is the same thing.
   Sum rule (Section 5.3). Basics: derivative rules

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