Prove the following
Question:

$\tan ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right), \frac{-1}{\sqrt{3}}<\frac{x}{a}<\frac{1}{\sqrt{3}}$

Solution:

Let $y=\tan ^{-1}\left[\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right]$

Put $x=a \tan \theta \quad \therefore \theta=\tan ^{-1} \frac{x}{a}$

$y=\tan ^{-1}\left[\frac{3 a^{2} \cdot a \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a \cdot a^{2} \tan ^{2} \theta}\right]$

$y=\tan ^{-1}\left[\frac{3 a^{3} \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a^{3} \tan ^{2} \theta}\right]$

$y=\tan ^{-1}\left[\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right]$

$y=\tan ^{-1}[\tan 3 \theta] \quad\left[\because \tan 3 \theta=\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right]$

$\Rightarrow y=3 \theta \Rightarrow y=3 \tan ^{-1} \frac{x}{a}$

Differentiating both sides w.r.t. $x$

$\frac{d y}{d x}=3 \cdot \frac{d}{d x}\left(\tan ^{-1} \frac{x}{a}\right)$

$=3 \cdot \frac{1}{1+\frac{x^{2}}{a^{2}}} \cdot \frac{d}{d x} \cdot\left(\frac{x}{a}\right)=3 \cdot \frac{a^{2}}{a^{2}+x^{2}} \cdot \frac{1}{a}=\frac{3 a}{a^{2}+x^{2}}$

Thus, $\frac{d y}{d x}=\frac{3 a}{a^{2}+x^{2}}$.

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