If $x=\sin ^{-1}\left(3 t-4 t^3\right)$ and $y=\cos ^{-1}\left(\sqrt{1-t^2}\right)$, then $\frac{d y}{d x}$ is equal to
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If $x=\sin ^{-1}\left(3 t-4 t^3\right)$ and $y=\cos ^{-1}\left(\sqrt{1-t^2}\right)$, then $\frac{d y}{d x}$ is equal to

(A) $1 / 2$

(B) $2 / 5$

(C) $3 / 2$

(D) $1 / 3$

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Best answer

SOLUTION —

Given that, $y=\cos ^{-1} \sqrt{1-t^2}=\sin ^{-1} t$

and $x=\sin ^{-1}\left(3 t-4 t^3\right)=3 \sin ^{-1} t$

$\therefore \quad \frac{d y}{d t}=\frac{1}{\sqrt{1-t^2}}$

and $\frac{d x}{d t}=\frac{3}{\sqrt{1-t^2}}$

$\therefore \quad \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\left(\frac{1}{\sqrt{1-t^2}}\right)}{3\left(\frac{1}{\sqrt{1-t^2}}\right)}=\frac{1}{3}$

So, The correct option will be (D).

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