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If $y=\sin ^{-1}\left[x \sqrt{1-x}-\sqrt{x} \cdot \sqrt{1-x^2}\right]$ and $0 < x < 1$, then $\frac{d y}{d x}$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{1-x^2}}-\frac{1}{2 \sqrt{x-x^2}}$
We have,
$$
\begin{aligned}
\text { ve, } y & =\sin ^{-1}\left(x \sqrt{1-x}-\sqrt{x} \sqrt{1-x^2}\right) \\
y & =\sin ^{-1} x-\sin ^{-1} \sqrt{x} \\
\frac{d y}{d x} & =\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x}} \times \frac{1}{2 \sqrt{x}} \\
\frac{d y}{d x} & =\frac{1}{\sqrt{1-x^2}}-\frac{1}{2 \sqrt{x-x^2}}
\end{aligned}
$$
$$
\begin{aligned}
\text { ve, } y & =\sin ^{-1}\left(x \sqrt{1-x}-\sqrt{x} \sqrt{1-x^2}\right) \\
y & =\sin ^{-1} x-\sin ^{-1} \sqrt{x} \\
\frac{d y}{d x} & =\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x}} \times \frac{1}{2 \sqrt{x}} \\
\frac{d y}{d x} & =\frac{1}{\sqrt{1-x^2}}-\frac{1}{2 \sqrt{x-x^2}}
\end{aligned}
$$
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