Example 24: Derivative of $\sin^{-1}x$ is $\frac{1}{\sqrt{1-x^2}}$

Basics you need

Solution

1. Rewrite the inverse as an equation we can differentiate

   Let $y = \sin^{-1}x$ with $x\in[-1,1]$⁠. By the meaning of $\sin^{-1}$⁠,

   $$\sin y = x \qquad\text{and}\qquad -\frac{\pi}{2} \le y \le \frac{\pi}{2}.$$
   We have a derivative for $\sin$ but not yet for $\sin^{-1}$⁠. The equation $\sin y = x$ contains only the function we know. Keeping the range condition next to it is essential: $\sin y = x$ alone has infinitely many solutions $y$⁠, and the range picks out the one that equals $\sin^{-1}x$⁠. Step 4 will need it.
   $\sin(\sin^{-1}x) = x$ for $x\in[-1,1]$ (Chapter 2, Section 2.3); principal value branch $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (Chapter 2, Section 2.2).

2. Differentiate both sides with respect to $x$

   $\sin y = x$ holds for every $x$ in $[-1,1]$⁠, so the two sides have the same derivative. Assuming $\frac{dy}{dx}$ exists (as the question allows), the chain rule with outer function $\sin$ and inner function $y(x)$ gives

   $$\cos y\cdot\frac{dy}{dx} = 1.$$
   The left side is a composite: sine of a quantity $y$ that depends on $x$⁠. Its derivative is $\cos y$ times the rate $\frac{dy}{dx}$ of the inner quantity. The right side is $x$⁠, whose derivative is $1$⁠.
   Implicit differentiation (Section 5.3.2) with the chain rule (Theorem 4). Lesson: implicit differentiation

3. Solve for $\frac{dy}{dx}$⁠, and see where this is possible

   $$\begin{aligned} \frac{dy}{dx} &= \frac{1}{\cos y} = \frac{1}{\cos(\sin^{-1}x)}, \\ &\qquad \cos y \ne 0. \end{aligned}$$

   In the range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠, $\cos y = 0$ only at $y = \pm\frac{\pi}{2}$⁠, that is, at $\sin^{-1}x = \pm\frac{\pi}{2}$⁠, that is, at $x = \pm1$⁠. So the formula is valid for

   $$x \in (-1, 1).$$
   We divide by $\cos y$⁠, which needs $\cos y \ne 0$⁠. Translating the condition back to $x$⁠: $y = \frac{\pi}{2}$ corresponds to $x = \sin\frac{\pi}{2} = 1$ and $y = -\frac{\pi}{2}$ to $x = -1$⁠. At those two points the equation $\cos y\cdot\frac{dy}{dx} = 1$ would read $0 = 1$⁠: no finite derivative can satisfy it.
   $\cos\left(\pm\frac{\pi}{2}\right) = 0$⁠, and $\cos y > 0$ strictly between them. Basics: trigonometric identities

4. Write $\cos y$ in terms of $x$⁠, choosing the correct sign

   For $x\in(-1,1)$⁠, the Pythagorean identity and $\sin y = x$ give

   $$\cos^2 y = 1 - \sin^2 y = 1 - x^2.$$

   So $\cos y = \sqrt{1-x^2}$ or $\cos y = -\sqrt{1-x^2}$⁠. Which one? Here $-\frac{\pi}{2} < y < \frac{\pi}{2}$⁠, and cosine is positive on that interval, so

   $$\cos y = +\sqrt{1 - x^2}.$$
   Squaring loses the sign, so after taking a square root we must decide it with extra information. That information is the principal range from Step 1. Had we used a branch of the inverse with $y$ in $\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$⁠, cosine would be negative there and the minus sign would be correct instead. The sign is also visible on the graph: $\sin^{-1}x$ increases, so its slope must be positive.
   $\sqrt{a}$ always means the non-negative root, so $\cos^2 y = 1-x^2$ gives $\cos y = \pm\sqrt{1-x^2}$ and the sign has to be chosen. Basics: trigonometric identities

5. Conclude, and look at the endpoints

   $$\begin{gathered} \frac{d}{dx}\left(\sin^{-1}x\right) = \frac{1}{\sqrt{1-x^2}}, \\ -1 < x < 1. \end{gathered}$$

   As $x$ approaches $\pm 1$ from inside, $\sqrt{1-x^2}$ approaches $0$ and the slope grows without bound: the graph becomes vertical at the ends.

   The derivative does not exist at $x = \pm1$⁠. For example, at $x = 1$ the difference quotient $\frac{\sin^{-1}(1-h) - \frac{\pi}{2}}{-h}$ (for small $h > 0$⁠) grows without limit: with $h = 0.0001$ it is already about $141$⁠. This is why the book's table gives the domain of $f'$ as $(-1, 1)$⁠, smaller than the domain $[-1, 1]$ of $f$⁠.
   Table after Example 24 (Section 5.3.3): $\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}$⁠, $\frac{d}{dx}\cos^{-1}x = \frac{-1}{\sqrt{1-x^2}}$⁠, both on $(-1,1)$⁠, and $\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$ on $\R$⁠. Lesson: derivatives of inverse trig functions

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