Exercise 5.4 · Q2: Differentiate $e^{\sin^{-1}x}$

Basics you need

Solution

1. Where is the function defined?

   The inner function $\sin^{-1}x$ needs $-1 \le x \le 1$⁠. Its value $t$ then lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠, and $e^t$ is defined for every real $t$⁠. So the domain of $y$ is the closed interval

   $$-1 \le x \le 1.$$

   Since $\sin^{-1}x$ increases with $x$ and $e^t$ increases with $t$⁠, $y$ runs from $e^{-\pi/2} \approx 0.2079$ at $x = -1$ up to $e^{\pi/2} \approx 4.8105$ at $x = 1$⁠.

   The exponential accepts any input, so it adds no restriction; the whole restriction comes from the inner function. Knowing the domain is closed matters in Step 4: the end points belong to the domain, so we must say what happens to the derivative there.
   Domain $[-1, 1]$ and principal range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ of $\sin^{-1}$⁠, Chapter 2. Lesson: derivatives of inverse trig functions

2. Name the inner and outer functions, and differentiate each

   Inner: $t = \sin^{-1}x$⁠. Outer: $y = e^t$⁠.

   $$\begin{aligned} \frac{dt}{dx} &= \frac{1}{\sqrt{1-x^2}}, \quad -1 < x < 1, \\ \frac{dy}{dt} &= e^t, \quad \text{all real } t. \end{aligned}$$
   To compute $e^{\sin^{-1}x}$ you first find the angle $\sin^{-1}x$⁠, then raise $e$ to it: the job done first is the inner function. The inner derivative comes from Example 24; it needs $1 - x^2 > 0$⁠, which excludes the end points $x = \pm1$⁠.
   Derivative of $\sin^{-1}x$⁠: Example 24 and the table after it, Section 5.3.3. Derivative of $e^t$⁠: Theorem 5 (1), Section 5.4. Lesson: exponential and logarithmic functions

3. Multiply (chain rule)

   For $-1 < x < 1$⁠, both link derivatives exist, so

   $$\begin{aligned} \frac{dy}{dx} &= \frac{dy}{dt}\cdot\frac{dt}{dx} = e^t\cdot\frac{1}{\sqrt{1-x^2}} \\ &= \frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}. \end{aligned}$$
   The outer derivative $e^t$ must be written back in terms of $x$⁠: with $t = \sin^{-1}x$ it is $e^{\sin^{-1}x}$⁠, the original function itself. So the derivative is "the function, divided by $\sqrt{1-x^2}$". Both factors are positive, so the slope is positive everywhere on $(-1, 1)$⁠, and the graph rises (Chapter 6, Section 6.3 turns "positive derivative" into "increasing function").
   Chain rule, Theorem 4, Section 5.3.1. Lesson: the chain rule

4. What happens at the end points $x = \pm 1$⁠?

   At $x = \pm1$ the formula has denominator $\sqrt{1 - 1} = 0$⁠, so it gives no value there. A formula failing is not yet a proof that no derivative exists, so we test the difference quotient itself. At an end point only one side is available: from the left at $x = 1$⁠, from the right at $x = -1$⁠.

   At $x = 1$⁠, write the nearby point as $x = \cos\varphi$ with $0 < \varphi < \frac{\pi}{2}$⁠. Then $\sin^{-1}x = \frac{\pi}{2} - \varphi$⁠, the step is $h = 1 - \cos\varphi > 0$⁠, and $h \to 0^+$ exactly when $\varphi \to 0^+$⁠. The left-hand difference quotient is

   $$\begin{aligned} \frac{f(1) - f(1-h)}{h} &= \frac{e^{\pi/2} - e^{\pi/2-\varphi}}{1 - \cos\varphi} \\ &= \frac{e^{\pi/2}\left(1 - e^{-\varphi}\right)}{1 - \cos\varphi}. \end{aligned}$$

   Two inequalities bound it from below:

   • bottom: $1 - \cos\varphi = 2\sin^2\frac{\varphi}{2} \le \frac{\varphi^2}{2}$⁠, because $\sin t < t$ for $0 < t < \frac{\pi}{2}$⁠;
   • top: $e^{\varphi} \ge 1 + \varphi$⁠, so $e^{-\varphi} \le \frac{1}{1+\varphi}$ and $1 - e^{-\varphi} \ge \frac{\varphi}{1+\varphi}$⁠.
   $$\frac{f(1) - f(1-h)}{h} \ge \frac{2e^{\pi/2}}{\varphi(1+\varphi)},$$

   which grows without bound as $\varphi \to 0^+$⁠. At $x = -1$⁠, write $x = -\cos\varphi$⁠: then $\sin^{-1}x = -\frac{\pi}{2} + \varphi$⁠, and the same two inequalities give

   $$\begin{aligned} \frac{f(-1+h) - f(-1)}{h} &= \frac{e^{-\pi/2}\left(e^{\varphi} - 1\right)}{1 - \cos\varphi} \\ &\ge \frac{2e^{-\pi/2}}{\varphi}, \end{aligned}$$

   which also grows without bound. So the function is defined at $\pm1$ but is not differentiable there.

   A derivative must be a finite limit of the difference quotient (Section 5.3); at an end point of the domain the book takes the one-sided limit from inside. Here both one-sided quotients exceed any number you name once $\varphi$ is small enough, so no finite limit exists. Writing the nearby point as $\cos\varphi$ turns $\sin^{-1}x$ into the simple angle $\frac{\pi}{2} - \varphi$⁠, which is why it is the convenient substitution. The two inequalities are used here, not proved: $\sin t < t$ comes from Class XI, and $e^{s} \ge 1 + s$ is proved with Chapter 6 (see the table at the end).
   Differentiable at a point means the difference quotient has a finite limit (Section 5.3). Lesson: differentiability

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