Exercise 5.3 · Q9: $\frac{dy}{dx}$ if $y=\sin^{-1}\frac{2x}{1+x^2}$

Basics you need

Solution

1. $y$ is defined for every real $x$

   $\sin^{-1}$ needs its input in $[-1, 1]$⁠, so we need $|2x| \le 1 + x^2$⁠. This always holds:

   $$1 + x^2 - 2|x| = (1 - |x|)^2 \ge 0,$$

   with equality only when $|x| = 1$⁠. So $\left|\frac{2x}{1 + x^2}\right| \le 1$ for all $x$⁠, and it equals $1$ only at $x = \pm1$⁠.

   This tells us the domain is all of $\mathbf{R}$⁠, so the question really is about every real $x$⁠. It also flags $x = \pm1$⁠: there the input of $\sin^{-1}$ is $\pm1$⁠, where $\sin^{-1}$ itself has no derivative.
   $x^2 = |x|^2$ and $(a - b)^2 \ge 0$⁠.

2. Substitute $x = \tan\theta$

   Let $\theta = \tan^{-1}x$⁠. This is defined for every real $x$⁠, gives $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠, and $\tan\theta = x$⁠. Since $\cos\theta \ne 0$ on this interval,

   $$\begin{aligned} \frac{2x}{1 + x^2} &= \frac{2\tan\theta}{1 + \tan^2\theta} = \frac{2\tan\theta}{\sec^2\theta} \\ &= 2\,\frac{\sin\theta}{\cos\theta}\cos^2\theta = \sin 2\theta. \end{aligned}$$

   So $y = \sin^{-1}(\sin 2\theta)$⁠, where $2\theta \in (-\pi, \pi)$⁠.

   The substitution is legitimate for every $x$ because $\tan$ takes every real value on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠. But $2\theta$ can be anywhere in $(-\pi, \pi)$⁠, which is wider than the principal range of $\sin^{-1}$⁠. So the next step must split into cases.
   $1 + \tan^2\theta = \sec^2\theta$⁠, $2\sin\theta\cos\theta = \sin 2\theta$ (Class XI). Basics: trigonometric identities

3. Case $-1 \le x \le 1$⁠: $y = 2\tan^{-1}x$

   $\tan^{-1}$ is increasing, with $\tan^{-1}(-1) = -\frac{\pi}{4}$ and $\tan^{-1}1 = \frac{\pi}{4}$⁠. So

   $$-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}, \quad\text{so}\quad -\frac{\pi}{2} \le 2\theta \le \frac{\pi}{2}.$$

   Now $2\theta$ is in the principal range of $\sin^{-1}$⁠, so $\sin^{-1}(\sin 2\theta) = 2\theta$⁠:

   $$y = 2\theta = 2\tan^{-1}x \quad (-1 \le x \le 1).$$
   This is the case the textbook's answer describes. The range check is what allows $\sin^{-1}$ and $\sin$ to cancel.
   $\sin^{-1}(\sin\phi) = \phi$ for $\phi \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (Chapter 2, Section 2.3). Basics: composite functions

4. Cases $x > 1$ and $x < -1$

   $x > 1$⁠. Then $\frac{\pi}{4} < \theta < \frac{\pi}{2}$ and $\frac{\pi}{2} < 2\theta < \pi$⁠, outside the principal range. Use $\sin 2\theta = \sin(\pi - 2\theta)$⁠, where $0 < \pi - 2\theta < \frac{\pi}{2}$ is inside it:

   $$y = \pi - 2\theta = \pi - 2\tan^{-1}x \quad (x > 1).$$

   $x < -1$⁠. Then $-\frac{\pi}{2} < \theta < -\frac{\pi}{4}$ and $-\pi < 2\theta < -\frac{\pi}{2}$⁠. Now $\sin(-\pi - 2\theta) = -\sin(\pi + 2\theta) = \sin 2\theta$⁠, and $-\frac{\pi}{2} < -\pi - 2\theta < 0$ is inside the range:

   $$y = -\pi - 2\theta = -\pi - 2\tan^{-1}x \quad (x < -1).$$

   At $x = 1$ both $2\tan^{-1}x$ and $\pi - 2\tan^{-1}x$ equal $\frac{\pi}{2}$⁠; at $x = -1$ both $2\tan^{-1}x$ and $-\pi - 2\tan^{-1}x$ equal $-\frac{\pi}{2}$⁠. So the pieces join up and $y$ is continuous.

   To undo $\sin$ we need an angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with the same sine as $2\theta$⁠. Reflecting through $\pm\frac{\pi}{2}$ ($\phi \mapsto \pi - \phi$⁠, or $\phi \mapsto -\pi - \phi$⁠) keeps the sine and brings the angle into range.
   $\sin(\pi - \phi) = \sin\phi$⁠, $\sin(\pi + \phi) = -\sin\phi$ (Class XI). Basics: trigonometric identities

5. Differentiate each piece

   Using $\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$⁠, and that the constants $\pm\pi$ have derivative $0$⁠:

   $$\begin{aligned} -1 < x < 1:&\quad \frac{dy}{dx} = \frac{2}{1 + x^2}, \\ |x| > 1:&\quad \frac{dy}{dx} = -\frac{2}{1 + x^2}. \end{aligned}$$
   On each open interval $y$ agrees with a simple formula, so its derivative there is the derivative of that formula. The endpoints $\pm1$ need separate treatment, because there the formula changes.
   $\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$ (table after Example 24, Section 5.3.3); constant multiple and sum rules. Lesson: derivatives of inverse trig functions

6. At $x = \pm1$⁠: one-sided derivatives differ

   At $x = 1$⁠, with $y(1) = \frac{\pi}{2}$⁠:

   • for $h < 0$ small, $1 + h \in (-1, 1)$⁠, so $y = 2\tan^{-1}x$ at both points of the quotient $\frac{y(1+h) - y(1)}{h}$⁠. The quotient is therefore the difference quotient of the differentiable function $2\tan^{-1}x$ at $1$⁠, and its limit as $h \to 0^-$ is that function's derivative at $1$⁠: the left-hand derivative is $\frac{2}{1 + 1^2} = 1$⁠;
   • for $h > 0$⁠, $y = \pi - 2\tan^{-1}x$ at both points (Step 4 showed it also gives $y(1)$⁠), so the quotient is the difference quotient of $\pi - 2\tan^{-1}x$⁠, and its limit as $h \to 0^+$⁠, the right-hand derivative, is $-\frac{2}{1 + 1^2} = -1$⁠.

   $1 \ne -1$⁠, so $y$ is not differentiable at $x = 1$⁠. In the same way, at $x = -1$ the left-hand derivative is $-1$ and the right-hand derivative is $1$⁠, so $y$ is not differentiable at $x = -1$ either.

   The graph has corners at $(1, \frac{\pi}{2})$ and $(-1, -\frac{\pi}{2})$⁠, like $|x|$ at $0$ (see the figure). Differentiability needs the two one-sided limits of the difference quotient to be equal, and here they are not.
   Left-hand and right-hand derivatives (Section 5.3; compare $|x|$ at $0$ after Theorem 3). Lesson: differentiability

7. Cross-check with the chain rule directly

   Let $u = \frac{2x}{1 + x^2}$⁠. By the quotient rule,

   $$u' = \frac{2(1 + x^2) - 2x\cdot2x}{(1 + x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2}.$$

   Also

   $$1 - u^2 = \frac{(1 + x^2)^2 - 4x^2}{(1 + x^2)^2} = \frac{(1 - x^2)^2}{(1 + x^2)^2},$$

   so $\sqrt{1 - u^2} = \frac{|1 - x^2|}{1 + x^2}$⁠. For $x \ne \pm1$ (so that $|u| < 1$⁠),

   $$\begin{aligned} \frac{dy}{dx} &= \frac{1}{\sqrt{1 - u^2}}\cdot u' \\ &= \frac{2(1 - x^2)}{(1 + x^2)\,|1 - x^2|}. \end{aligned}$$

   The fraction $\frac{1 - x^2}{|1 - x^2|}$ is $1$ when $|x| < 1$ and $-1$ when $|x| > 1$⁠, so this is exactly Step 5.

   The square root of a square is an absolute value: $\sqrt{(1 - x^2)^2} = |1 - x^2|$⁠, not $1 - x^2$⁠. Forgetting this is how the sign change for $|x| > 1$ gets lost.
   $\frac{d}{du}\sin^{-1}u = \frac{1}{\sqrt{1 - u^2}}$ for $-1 < u < 1$ (Example 24); chain rule, Theorem 4; quotient rule (3), Section 5.3. Basics: derivative rules

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