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…
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
- $\frac{1}{2 \sqrt{1-x^2}}-\frac{1}{x \sqrt{1-x^2}}$
- $\frac{1}{\sqrt{1-x^2}}-\frac{1}{2 \sqrt{x-x^2}}$
- $\frac{1}{2 \sqrt{x-x^2}}+\frac{1}{\sqrt{1-x^2}}$
- $\frac{-1}{\sqrt{1-x^2}}-\frac{1}{x \sqrt{1-x^2}}$
Solution
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}
$
Asked in: AP EAMCET 2021 (25 Aug Shift 1)
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