Miscellaneous · Q4: Differentiate $\sin^{-1}(x\sqrt x)$ on $[0,1]$

Basics you need

Solution

1. Where the inner function lands

   Let $u = x\sqrt x$⁠. For $0 \le x \le 1$⁠, $\sqrt x$ is defined and $0 \le \sqrt x \le 1$⁠, so

   $$0 \le u = x\sqrt x \le 1\cdot 1 = 1,$$

   with $u = 1$ only when $x = 1$⁠. So $\sin^{-1}u$ is defined on the whole interval, and $u \in [0, 1)$ for $0 \le x < 1$⁠.

   $\sin^{-1}$ accepts only inputs in $[-1,1]$⁠, and its derivative formula only inputs in $(-1,1)$⁠. Knowing that $u$ reaches 1 exactly at $x = 1$ tells us in advance that $x = 1$ needs separate treatment.
   $\sin^{-1}: [-1,1] \to \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ (Chapter 2, Section 2.2). Basics: composite functions

2. Differentiate the inner function $u = x\sqrt x$

   For $x > 0$⁠, by the product rule with factors $x$ and $\sqrt x$⁠:

   $$\begin{aligned} \frac{du}{dx} &= 1\cdot\sqrt x + x\cdot\frac{1}{2\sqrt x} \\ &= \sqrt x + \frac{\sqrt x}{2} = \frac{3}{2}\sqrt x. \end{aligned}$$
   We used $\frac{x}{\sqrt x} = \sqrt x$ for $x > 0$⁠. The same result comes from writing $u = x^{3/2}$ and using $\frac{d}{dx}x^n = nx^{n-1}$ with $n = \frac32$⁠. The product rule route avoids any doubt about fractional powers.
   Product rule (2), Section 5.3; $\frac{d}{dx}\sqrt x = \frac{1}{2\sqrt x}$ for $x > 0$ (Class XI). Basics: derivative rules

3. Chain rule for $0 < x < 1$

   Here $u \in (0,1)$⁠, so the formula for $\sin^{-1}$ applies. Also $u^2 = x^2\cdot x = x^3$⁠. Hence

   $$\begin{aligned} \frac{dy}{dx} &= \frac{1}{\sqrt{1-u^2}}\cdot\frac{du}{dx} \\ &= \frac{1}{\sqrt{1-x^3}}\cdot\frac{3}{2}\sqrt x = \frac{3\sqrt x}{2\sqrt{1-x^3}}. \end{aligned}$$

   Since $x > 0$ and $1 - x^3 > 0$⁠, the two square roots combine into one:

   $$\frac{dy}{dx} = \frac{3}{2}\sqrt{\frac{x}{1-x^3}}.$$
   $\frac{\sqrt a}{\sqrt b} = \sqrt{\frac ab}$ holds when $a \ge 0$ and $b > 0$⁠, which is the case here. This puts the answer in exactly the form printed in the answer key.
   Chain rule, Theorem 4, Section 5.3.1; $\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}$ on $(-1,1)$⁠, table after Example 24, Section 5.3.3. Lesson: chain rule

4. The end point $x = 0$⁠: a right-hand derivative

   The domain starts at 0, so only $h > 0$ is allowed. With $y(0) = \sin^{-1}0 = 0$ and $s = h\sqrt h$⁠:

   $$\begin{aligned} \frac{y(h) - y(0)}{h} &= \frac{\sin^{-1}s}{s}\cdot\frac{s}{h} \\ &= \frac{\sin^{-1}s}{s}\cdot\sqrt h. \end{aligned}$$

   As $h \to 0^+$⁠, $s \to 0^+$ and $\frac{\sin^{-1}s}{s} \to 1$ (put $\theta = \sin^{-1}s$⁠; this is $\frac{\theta}{\sin\theta} \to 1$⁠), while $\sqrt h \to 0$⁠. So the right-hand derivative at 0 is $1\cdot 0 = 0$⁠, which is also what the formula gives at $x = 0$⁠.

   Step 2 divided by $\sqrt x$⁠, so it does not cover $x = 0$⁠. At an end point the book takes the one-sided limit (Section 5.3), and here it exists and agrees with the formula. So the formula is valid on $0 \le x < 1$⁠.
   $\lim_{\theta\to 0}\frac{\sin\theta}{\theta} = 1$ (Class XI, Limits). One-sided derivatives at end points, Section 5.3. Basics: left-hand and right-hand limits

5. The end point $x = 1$⁠: no derivative

   Here $y(1) = \sin^{-1}1 = \frac{\pi}{2}$⁠, and only $h > 0$ with $1 - h$ in the domain is allowed. Let $u = (1-h)\sqrt{1-h} \in [0,1)$ and $\varphi = \frac{\pi}{2} - \sin^{-1}u$⁠, an angle in $\left(0, \frac{\pi}{2}\right]$⁠. Then $\sin\varphi = \cos(\sin^{-1}u) = \sqrt{1-u^2}$⁠, and $u^2 = (1-h)^3$⁠. Because $\varphi \ge \sin\varphi$ for such angles,

   $$\begin{aligned} \frac{y(1) - y(1-h)}{h} = \frac{\varphi}{h} &\ge \frac{\sqrt{1-(1-h)^3}}{h} \\ &= \frac{\sqrt{3 - 3h + h^2}}{\sqrt h}. \end{aligned}$$

   As $h \to 0^+$ the numerator tends to $\sqrt 3$ and the denominator to 0, so the quotient grows without bound. The left-hand derivative at $x = 1$ is not finite: $y$ is not differentiable at $x = 1$. The graph has a vertical tangent there.

   The formula $\frac32\sqrt{\frac{x}{1-x^3}}$ has a zero denominator at $x = 1$⁠, which already warns us. The inequality makes the warning precise. The expansion $1 - (1-h)^3 = 3h - 3h^2 + h^3$ is the binomial cube, and we divided top and bottom by $\sqrt h$⁠.
   $\cos(\sin^{-1}u) = \sqrt{1-u^2}$ for $u \in [-1,1]$ (cosine is $\ge 0$ on the principal range, as in Example 24). $\sin\varphi \le \varphi$ for $\varphi \ge 0$ (Class XI). Basics: trigonometric identities

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