Exercise 5.3 · Q11: $\frac{dy}{dx}$ if $y=\cos^{-1}\frac{1-x^2}{1+x^2}$

Basics you need

Solution

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

   Let $\theta = \tan^{-1}x$⁠, so $x = \tan\theta$ (possible for every real $x$⁠). Applying the increasing function $\tan^{-1}$ to $0 < x < 1$⁠:

   $$0 < \theta < \frac{\pi}{4}, \quad\text{so}\quad 0 < 2\theta < \frac{\pi}{2}.$$
   We will need the range of $2\theta$⁠, because $2\theta$ is the angle that ends up inside $\cos^{-1}(\cos\cdot)$⁠.
   $\tan^{-1}0 = 0$⁠, $\tan^{-1}1 = \frac{\pi}{4}$ (Chapter 2, Fig 2.5 (ii)). Basics: trigonometric identities

2. The fraction is $\cos 2\theta$

   Using $1 + \tan^2\theta = \sec^2\theta$⁠, that is, multiplying top and bottom by $\cos^2\theta \ne 0$⁠:

   $$\begin{aligned} \frac{1 - x^2}{1 + x^2} &= \frac{1 - \tan^2\theta}{1 + \tan^2\theta} \\ &= \cos^2\theta - \sin^2\theta = \cos 2\theta. \end{aligned}$$
   This is the "cosine of a double angle in terms of tangent" formula. The denominator $1 + x^2$ is never zero, so the fraction is defined for every $x$⁠.
   $\cos 2\theta = \cos^2\theta - \sin^2\theta$ (Class XI). Basics: trigonometric identities

3. Simplify $y$⁠, checking the range

   Since $2\theta \in \left(0, \frac{\pi}{2}\right) \subset [0, \pi]$⁠,

   $$y = \cos^{-1}(\cos 2\theta) = 2\theta = 2\tan^{-1}x.$$
   $\cos^{-1}$ returns the unique angle in $[0, \pi]$ with the given cosine. $2\theta$ is such an angle, so it is the one returned.
   $\cos^{-1}(\cos\phi) = \phi$ for $\phi \in [0, \pi]$ (Chapter 2, Section 2.3). Basics: composite functions

4. Differentiate

   $$\frac{dy}{dx} = 2\cdot\frac{1}{1 + x^2} = \frac{2}{1 + x^2}.$$
   $\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$ (table after Example 24, Section 5.3.3). Lesson: derivatives of inverse trig functions

5. Cross-check with the chain rule directly

   Let $u = \frac{1 - x^2}{1 + x^2}$⁠. The quotient rule gives

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

   Also

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

   so $\sqrt{1 - u^2} = \frac{2|x|}{1 + x^2}$⁠. 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 $0 < x < 1$⁠, $\frac{x}{|x|} = 1$⁠, and this is $\frac{2}{1 + x^2}$⁠, as in Step 4.

   The same calculation shows what happens outside the interval. For every $x > 0$ the answer is $\frac{2}{1 + x^2}$ (indeed $2\theta \in (0, \pi)$ there), so the condition $x < 1$ is not needed. For $x < 0$ the answer is $-\frac{2}{1 + x^2}$⁠, and at $x = 0$ the one-sided derivatives $\pm2$ differ.
   Chain rule, Theorem 4; quotient rule (3), Section 5.3; $\sqrt{a^2} = |a|$⁠. Basics: derivative rules

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