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

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(D) $y=\sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}$Differentiating with respect to $x$,we get$\frac{dy}{dx} = \frac{d}{dx}\left(\sqrt{\frac{1-\sin ^{

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$-1$

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(D) $y=\sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}$Differentiating with respect to $x$,we get$\frac{dy}{dx} = \frac{d}{dx}\left(\sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}\right)$$= \frac{1}{2 \sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}} \cdot \frac{d}{dx}\left(\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}\right)$$= \frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \cdot \left[ \frac{(1+\sin ^{-1} x) \cdot \frac{d}{dx}(1-\sin ^{-1} x) - (1-\sin ^{-1} x) \cdot \frac{d}{dx}(1+\sin ^{-1} x)}{(1+\sin ^{-1} x)^2} \right]$$= \frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \cdot \left[ \frac{(1+\sin ^{-1} x) \cdot \frac{-1}{\sqrt{1-x^2}} - (1-\sin ^{-1} x) \cdot \frac{1}{\sqrt{1-x^2}}}{(1+\sin ^{-1} x)^2} \right]$$= \frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \cdot \frac{1}{\sqrt{1-x^2}} \cdot \left[ \frac{-1-\sin ^{-1} x - 1 + \sin ^{-1} x}{(1+\sin ^{-1} x)^2} \right]$$= \frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \cdot \frac{1}{\sqrt{1-x^2}} \cdot \frac{-2}{(1+\sin ^{-1} x)^2}$$= \frac{-\sqrt{1+\sin ^{-1} x}}{\sqrt{1-\sin ^{-1} x} \cdot \sqrt{1-x^2} \cdot (1+\sin ^{-1} x)^2}$At $x=0$,$\sin^{-1}(0) = 0$:$\left(\frac{dy}{dx}\right)_{x=0} = \frac{-\sqrt{1+0}}{\sqrt{1-0} \cdot \sqrt{1-0} \cdot (1+0)^2} = \frac{-1}{1 \cdot 1 \cdot 1} = -1$

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

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