If y 1-x^2= ^ -1 x, then find d y d x - Vidyadip

Question

If $y \sqrt{1-x^2}=\sin ^{-1} x$, then find $\frac{d y}{d x}$

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Answer

Given $
y \sqrt{1-x^2}=\sin ^{-1} x
$
differentiating both sides w.r.t. $x$$
\begin{aligned}
\frac{d y}{d x} \sqrt{1-x^2}+y \frac{1}{2 \sqrt{1-x^2}} & \times(-2 x)=\frac{1}{\sqrt{1-x^2}} \\
\sqrt{1-x^2} \frac{d y}{d x}-\frac{x y}{\sqrt{1-x^2}} & =\frac{1}{\sqrt{1-x^2}} \\
\left(1-x^2\right) \frac{d y}{d x}-x y & =1 \\
\left(1-x^2\right) \frac{d y}{d x} & =1+x y \\
\frac{d y}{d x} & =\frac{1+x y}{\left(1-x^2\right)}
\end{aligned}
$

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