If y=sqrt{frac{1-sin ^{-1}(x)}{1+sin ^{-1}(x) | vedclass

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(B) Given $y=\sqrt{\frac{1-\sin ^{-1}(x)}{1+\sin ^{-1}(x)}}$.Taking the natural logarithm on both sides:$\log y = \frac{1}{2} [\log(1-\sin^{-1}x) - \l

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$-1$

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(B) Given $y=\sqrt{\frac{1-\sin ^{-1}(x)}{1+\sin ^{-1}(x)}}$.Taking the natural logarithm on both sides:$\log y = \frac{1}{2} [\log(1-\sin^{-1}x) - \log(1+\sin^{-1}x)]$.Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{1-\sin^{-1}x} \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right) - \frac{1}{1+\sin^{-1}x} \cdot \left(\frac{1}{\sqrt{1-x^2}}\right) \right]$.Simplifying the expression:$\frac{dy}{dx} = -\frac{y}{2\sqrt{1-x^2}} \left( \frac{1}{1-\sin^{-1}x} + \frac{1}{1+\sin^{-1}x} \right)$.At $x=0$,$\sin^{-1}(0)=0$ and $y=1$:$\left(\frac{dy}{dx}\right)_{(0,1)} = -\frac{1}{2\sqrt{1-0}} \left( \frac{1}{1-0} + \frac{1}{1+0} \right) = -\frac{1}{2} (1+1) = -1$.

If $y=\sqrt{\frac{1-\sin ^{-1}(x)}{1+\sin ^{-1}(x)}}$,then $\frac{dy}{dx}$ at $x=0$ and $y=1$ is

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