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If $y=\tan ^{-1}\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right), \frac{-1}{\sqrt{3}} < \frac{x}{a} < \frac{1}{\sqrt{3}}$, Then find $\frac{d y}{d x}$.
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Verified Answer
$$
\begin{aligned}
&\text {Put } x=a \tan \theta \Rightarrow \theta=\tan ^{-1} \frac{x}{a}\\
&\therefore \mathrm{y}=\tan ^{-1}\left[\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right]\\
&=\tan ^{-1}(\tan 3 \theta)=3 \theta \quad\left[\because \tan 3 \theta=\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right]\\
&=\tan ^{-1}(\tan 3 \theta)=3 \theta\\
&=3 \tan ^{-1} \frac{\mathrm{x}}{\mathrm{a}} \quad\left[\because \theta=\tan ^{-1} \frac{\mathrm{x}}{\mathrm{a}}\right]\\
&\therefore \quad \frac{d y}{d x}=3 \cdot \frac{d}{d x} \tan ^{-1} \frac{x}{a}=3 \cdot\left[\frac{1}{1+\frac{x^2}{a^2}}\right] \cdot \frac{d}{d x} \cdot\left(\frac{x}{a}\right)
\end{aligned}
$$
$=3 \cdot \frac{a^2}{a^2+x^2} \cdot \frac{1}{a}=\frac{3 a}{a^2+x^2}$
\begin{aligned}
&\text {Put } x=a \tan \theta \Rightarrow \theta=\tan ^{-1} \frac{x}{a}\\
&\therefore \mathrm{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[\because \tan 3 \theta=\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right]\\
&=\tan ^{-1}(\tan 3 \theta)=3 \theta\\
&=3 \tan ^{-1} \frac{\mathrm{x}}{\mathrm{a}} \quad\left[\because \theta=\tan ^{-1} \frac{\mathrm{x}}{\mathrm{a}}\right]\\
&\therefore \quad \frac{d y}{d x}=3 \cdot \frac{d}{d x} \tan ^{-1} \frac{x}{a}=3 \cdot\left[\frac{1}{1+\frac{x^2}{a^2}}\right] \cdot \frac{d}{d x} \cdot\left(\frac{x}{a}\right)
\end{aligned}
$$
$=3 \cdot \frac{a^2}{a^2+x^2} \cdot \frac{1}{a}=\frac{3 a}{a^2+x^2}$
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