Exercise 5.3 · Q12: $\frac{dy}{dx}$ if $y=\sin^{-1}\frac{1-x^2}{1+x^2}$

Basics you need

Solution

1. Substitute $x = \tan\theta$ and locate $\theta$

   Let $\theta = \tan^{-1}x$⁠, so $x = \tan\theta$⁠. From $0 < x < 1$ and the increasing function $\tan^{-1}$⁠,

   $$0 < \theta < \frac{\pi}{4}, \quad\text{so}\quad 0 < 2\theta < \frac{\pi}{2}.$$
   The substitution is legitimate for every real $x$⁠. We need the interval for $\theta$ to check the range in Step 3.
   $\tan^{-1}0 = 0$⁠, $\tan^{-1}1 = \frac{\pi}{4}$ (Chapter 2, Fig 2.5 (ii)). Basics: trigonometric identities

2. Write the fraction as a sine

   $$\begin{aligned} \frac{1 - x^2}{1 + x^2} &= \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos 2\theta \\ &= \sin\left(\frac{\pi}{2} - 2\theta\right). \end{aligned}$$
   The first equality multiplies top and bottom by $\cos^2\theta \ne 0$⁠, as in Q11. The second turns the cosine into a sine, because $\sin^{-1}$ can only undo a sine.
   $\cos 2\theta = \cos^2\theta - \sin^2\theta$⁠; $\sin\left(\frac{\pi}{2} - \alpha\right) = \cos\alpha$ (Class XI). Basics: trigonometric identities

3. Simplify $y$⁠, checking the range

   From $0 < 2\theta < \frac{\pi}{2}$⁠,

   $$0 < \frac{\pi}{2} - 2\theta < \frac{\pi}{2},$$

   which lies inside $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠. Therefore

   $$\begin{aligned} y &= \sin^{-1}\left(\sin\left(\frac{\pi}{2} - 2\theta\right)\right) \\ &= \frac{\pi}{2} - 2\theta = \frac{\pi}{2} - 2\tan^{-1}x. \end{aligned}$$
   Subtracting from $\frac{\pi}{2}$ reverses the inequalities: the largest $2\theta$ gives the smallest $\frac{\pi}{2} - 2\theta$⁠. The resulting interval $\left(0, \frac{\pi}{2}\right)$ is in the principal range, so $\sin^{-1}$ and $\sin$ 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. Differentiate

   $$\frac{dy}{dx} = 0 - 2\cdot\frac{1}{1 + x^2} = -\frac{2}{1 + x^2}.$$
   The constant $\frac{\pi}{2}$ has derivative $0$⁠; the rest is a constant multiple of $\tan^{-1}x$⁠.
   $\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$ (table after Example 24, Section 5.3.3). Lesson: derivatives of inverse trig functions

5. Cross-check with the chain rule directly

   With $u = \frac{1 - x^2}{1 + x^2}$⁠, Q11 found $u' = \frac{-4x}{(1 + x^2)^2}$ and $\sqrt{1 - u^2} = \frac{2|x|}{1 + x^2}$⁠. For $x \ne 0$⁠,

   $$\begin{aligned} \frac{dy}{dx} &= \frac{1}{\sqrt{1 - u^2}}\cdot u' \\ &= \frac{1 + x^2}{2|x|}\cdot\frac{-4x}{(1 + x^2)^2} = \frac{-2x}{|x|\,(1 + x^2)}. \end{aligned}$$

   For $0 < x < 1$ this is $-\frac{2}{1 + x^2}$⁠, as in Step 4.

   Again the result holds for every $x > 0$⁠, not only $x < 1$⁠; for $x < 0$ the sign flips to $+\frac{2}{1 + x^2}$⁠. Notice also that $y$ here is $\frac{\pi}{2}$ minus the $y$ of Q11, so the two derivatives are negatives of each other.
   Chain rule, Theorem 4; $\frac{d}{du}\sin^{-1}u = \frac{1}{\sqrt{1 - u^2}}$⁠, Example 24. Basics: derivative rules

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