Differentiate the function $\tan^{-1} \left( \frac{3a^2x - x^3}{a^3 - 3ax^2} \right), \quad \frac{-1}{\sqrt{3}} < \frac{x}{a} < \frac{1}{\sqrt{3}}$ with respect to $x$.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\frac{3a}{a^2 + x^2}$ ##
Let $y = \tan^{-1} \left( \frac{3a^2x - x^3}{a^3 - 3ax^2} \right) \quad \dots(i)$
On putting $x = a \tan \theta \Rightarrow \theta = \tan^{-1} \frac{x}{a}$ in Eq. (i), we get
$y = \tan^{-1} \left[ \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \right]$
$= \tan^{-1} (\tan 3\theta) = 3\theta \quad \left[ ∵\tan 3\theta = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \right]$
$= 3 \tan^{-1} \frac{x}{a} \quad \left[ ∵\theta = \tan^{-1} \frac{x}{a} \right]$
$∴\frac{dy}{dx} = 3 \cdot \frac{d}{dx} \tan^{-1} \frac{x}{a} = 3 \cdot \left[ \frac{1}{1 + \frac{x^2}{a^2}} \right] \cdot \frac{d}{dx} \left( \frac{x}{a} \right)$
$= 3 \cdot \frac{a^2}{a^2 + x^2} \cdot \frac{1}{a} = \frac{3a}{a^2 + x^2}$