Exercise 5.3 · Q10: $\frac{dy}{dx}$ if $y=\tan^{-1}\frac{3x-x^3}{1-3x^2}$

Basics you need

Solution

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

   Let $\theta = \tan^{-1}x$⁠, so $x = \tan\theta$⁠. This works for every real $x$⁠. Applying the increasing function $\tan^{-1}$ to $-\frac{1}{\sqrt3} < x < \frac{1}{\sqrt3}$⁠:

   $$-\frac{\pi}{6} < \theta < \frac{\pi}{6}, \quad\text{so}\quad -\frac{\pi}{2} < 3\theta < \frac{\pi}{2}.$$
   The answer will be $\tan^{-1}(\tan 3\theta)$⁠, and that cancels to $3\theta$ only if $3\theta$ is in the principal range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠. Tripling $\left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$ lands exactly on it; this is why the question uses the limits $\pm\frac{1}{\sqrt3}$⁠.
   $\tan\frac{\pi}{6} = \frac{1}{\sqrt3}$⁠; $\tan^{-1}$ is increasing (Chapter 2, Fig 2.5 (ii)). Basics: trigonometric identities

2. The fraction is $\tan 3\theta$

   With $t = \tan\theta$⁠, $\tan 2\theta = \frac{2t}{1 - t^2}$⁠, and then

   $$\begin{aligned} \tan 3\theta &= \frac{\tan 2\theta + \tan\theta}{1 - \tan 2\theta\tan\theta} \\ &= \frac{2t + t(1 - t^2)}{(1 - t^2) - 2t^2} \\ &= \frac{3t - t^3}{1 - 3t^2} \end{aligned}$$

   (the second line multiplies top and bottom by $1 - t^2$⁠). With $t = x$⁠, the fraction in the question is $\tan 3\theta$⁠.

   No denominator is zero: $x^2 < \frac13$ gives $1 - 3x^2 > 0$ and $1 - x^2 > 0$⁠, and $\theta$⁠, $2\theta$⁠, $3\theta$ all lie in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠, where $\tan$ is defined.

   Every formula has conditions. The double- and triple-angle formulas need $\tan\theta$⁠, $\tan 2\theta$ and $\tan 3\theta$ to exist, which the interval guarantees.
   $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}$ (Class XI). Basics: trigonometric identities

3. Simplify $y$

   $$\begin{aligned} y &= \tan^{-1}(\tan 3\theta) = 3\theta \\ &= 3\tan^{-1}x, \end{aligned}$$

   because $3\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (Step 1).

   $\tan^{-1}$ returns the unique angle in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with the given tangent; $3\theta$ is such an angle, so it is the one returned.
   $\tan^{-1}(\tan\phi) = \phi$ for $\phi \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (Chapter 2, Section 2.3). Basics: composite functions

4. Differentiate

   $$\frac{dy}{dx} = 3\cdot\frac{1}{1 + x^2} = \frac{3}{1 + x^2}.$$
   A constant multiple stays in front, and $\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$⁠.
   Table after Example 24, Section 5.3.3. Lesson: derivatives of inverse trig functions

5. Cross-check with the chain rule directly

   Let $u = \frac{3x - x^3}{1 - 3x^2}$⁠. By the quotient rule the numerator of $u'$ is

   $$\begin{aligned} &(3 - 3x^2)(1 - 3x^2) - (3x - x^3)(-6x) \\ &= 3 - 12x^2 + 9x^4 + 18x^2 - 6x^4 \\ &= 3 + 6x^2 + 3x^4 = 3(1 + x^2)^2, \end{aligned}$$

   so $u' = \frac{3(1 + x^2)^2}{(1 - 3x^2)^2}$⁠. Also

   $$\begin{aligned} 1 + u^2 &= \frac{(1 - 3x^2)^2 + (3x - x^3)^2}{(1 - 3x^2)^2} \\ &= \frac{(1 + x^2)^3}{(1 - 3x^2)^2}, \end{aligned}$$

   since both numerators expand to $1 + 3x^2 + 3x^4 + x^6$⁠. Hence

   $$\frac{dy}{dx} = \frac{u'}{1 + u^2} = \frac{3(1 + x^2)^2}{(1 + x^2)^3} = \frac{3}{1 + x^2}.$$
   The direct route needs no substitution and agrees with Step 4. It is longer, which is why the substitution is the better method.
   Chain rule, Theorem 4; quotient rule (3), Section 5.3. Basics: derivative rules

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