The derivative of $y=\sin ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{2}\right)$ is
The derivative of $y=\sin ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{2}\right)$ is
- $\frac{1}{\sqrt{1+x^2}}$
- $\frac{1}{\sqrt{1-x^2}}$
- $\frac{1}{2 \sqrt{1+x^2}}$
- $\frac{1}{2 \sqrt{1-x^2}}$
Solution
$y=\sin ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{2}\right)$
Put, $x=\cos 2 \theta \Rightarrow 2 \theta=\cos ^{-1} x$
$
\begin{aligned}
\theta & =\frac{1}{2} \cos ^{-1} x \\
& =\sin ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}{2}\right) \\
& =\sin ^{-1}\left(\frac{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}{2}\right) \\
& =\sin ^{-1}\left(\frac{\cos \theta-\sin \theta}{\sqrt{2}}\right) \\
& =\sin ^{-1}\left(\frac{1}{\sqrt{2}} \cos \theta-\sin \theta \frac{1}{\sqrt{2}}\right) \\
& =\sin ^{-1}\left(\sin \frac{\pi}{4} \cos \theta-\cos \frac{\pi}{4} \sin \theta\right) \\
& =\sin ^{-1}\left[\sin \left(\frac{\pi}{4}-\theta\right)\right]=\frac{\pi}{4}-\theta \\
y & =\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x
\end{aligned}
$
Now, differentiate w.r.t ' $x$ ', on both sides
$
\frac{d y}{d x}=0-\frac{1}{2} \cdot \frac{-1}{\sqrt{1-x^2}} \Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{1-x^2}}
$
Hence, option (4) is correct
Asked in: AP EAMCET 2020 (22 Sep Shift 2)
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