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If $y=\tan \left(3 \tan ^{-1} x\right)$, then$\left(1-3 x^2\right) \frac{d^2 y}{d x^2}-12 x \frac{d y}{d x}=$
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The correct answer is:
$6(y-x)$
$y=\tan \left(3 \tan ^{-1} x\right)$
Let $\tan ^{-1} x=\theta$
$\Rightarrow \quad \tan \theta=x$
$\therefore \quad y=\tan 3 \theta=\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}$
$=\frac{3 x-x^3}{1-3 x^2}$
$\Rightarrow\left(1-3 x^2\right) y=3 x-x^3$
On differentiating both sides w.r.t. $x$,
$\left(1-3 x^2\right) \frac{d y}{d x}+y(-6 x)=3-3 x^2$
Again, differentiating both sides w.r.t. $x$,
$\left(1-3 x^2\right) \frac{d^2 y}{d x^2}+(-6 x) \frac{d y}{d x}-6\left(x \frac{d y}{d x}+y\right)=-6 x$
$\Rightarrow \quad\left(1-3 x^2\right) \frac{d^2 y}{d x^2}-12 x \frac{d y}{d x}=6(y-x)$
Let $\tan ^{-1} x=\theta$
$\Rightarrow \quad \tan \theta=x$
$\therefore \quad y=\tan 3 \theta=\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}$
$=\frac{3 x-x^3}{1-3 x^2}$
$\Rightarrow\left(1-3 x^2\right) y=3 x-x^3$
On differentiating both sides w.r.t. $x$,
$\left(1-3 x^2\right) \frac{d y}{d x}+y(-6 x)=3-3 x^2$
Again, differentiating both sides w.r.t. $x$,
$\left(1-3 x^2\right) \frac{d^2 y}{d x^2}+(-6 x) \frac{d y}{d x}-6\left(x \frac{d y}{d x}+y\right)=-6 x$
$\Rightarrow \quad\left(1-3 x^2\right) \frac{d^2 y}{d x^2}-12 x \frac{d y}{d x}=6(y-x)$
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