If $y=\sqrt{\frac{1-\sin ^{-1}(x)}{1+\sin ^{-1}(x)}}$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=0$…

If $y=\sqrt{\frac{1-\sin ^{-1}(x)}{1+\sin ^{-1}(x)}}$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=0$ and $y=1$ is
  1. $-2$
  2. $-1$
  3. $1$
  4. $2$

Solution

$y=\sqrt{\frac{1-\sin ^{-1}(x)}{1+\sin ^{-1}(x)}}$ Taking log on both sides, we get $\log y=\frac{1}{2}\left[\log \left(1-\sin ^{-1} x\right)-\log \left(1+\sin ^{-1} x\right)\right]$ Differentiating w. r. t. $x$, we get $\begin{aligned} & \frac{1}{y} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{2}\left[\frac{1}{1-\sin ^{-1} x} \cdot \frac{-1}{\sqrt{1-x^2}}-\frac{1}{1+\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}\right] \\ & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-y}{2 \sqrt{1-x^2}}\left(\frac{1}{1-\sin ^{-1} x}+\frac{1}{1+\sin ^{-1} x}\right) \\ & \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(0,1)}=\frac{1}{2(1)}\left(\frac{1}{1-0}+\frac{1}{1+0}\right)=-1 \end{aligned}$

Asked in: MHT CET 2023 (11 May Shift 1)

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