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If $\sin y=\sin 3 t$ and $x=\sin t$, then $\frac{d y}{d x}=$
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The correct answer is:
$\frac{3}{\sqrt{1-x^2}}$
$\sin y=\sin 3 t$
$\begin{aligned} & t=3 t \\ & y=3 \sin ^{-1} x \\ & \frac{d y}{d x}=\frac{d}{d x} 3 \sin ^{-1} x \\ & =\frac{3}{\sqrt{1-x^2}}\end{aligned}$
$\begin{aligned} & t=3 t \\ & y=3 \sin ^{-1} x \\ & \frac{d y}{d x}=\frac{d}{d x} 3 \sin ^{-1} x \\ & =\frac{3}{\sqrt{1-x^2}}\end{aligned}$
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