If y=frac{2}{sqrt{a^2-b^2}} tan ^{-1}left[sqr | vedclass

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(B) We have,$y=\frac{2}{\sqrt{a^2-b^2}} 	an ^{-1}\left[\sqrt{\frac{a-b}{a+b}} 	an \frac{x}{2}ight]$.Differentiating with respect to $x$:$\frac{d y

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$\frac{b}{a^2}$

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(B) We have,$y=\frac{2}{\sqrt{a^2-b^2}} 	an ^{-1}\left[\sqrt{\frac{a-b}{a+b}} 	an \frac{x}{2}ight]$.Differentiating with respect to $x$:$\frac{d y}{d x}=\frac{2}{\sqrt{a^2-b^2}} \cdot \frac{1}{1+\left(\frac{a-b}{a+b}ight) 	an ^2 \frac{x}{2}} \cdot \sqrt{\frac{a-b}{a+b}} \cdot \sec ^2 \frac{x}{2} \cdot \frac{1}{2}$$= \frac{1}{\sqrt{a^2-b^2}} \cdot \frac{a+b}{(a+b) + (a-b) 	an ^2 \frac{x}{2}} \cdot \sqrt{\frac{a-b}{a+b}} \cdot \sec ^2 \frac{x}{2}$$= \frac{\sqrt{a+b}}{\sqrt{a-b}} \cdot \frac{1}{\sqrt{a+b} \cdot \sqrt{a-b}} \cdot \frac{\sec ^2 \frac{x}{2}}{a(1+	an ^2 \frac{x}{2}) + b(1-	an ^2 \frac{x}{2})}$$= \frac{\sec ^2 \frac{x}{2}}{(a+b \cos x) \sec ^2 \frac{x}{2}} = \frac{1}{a+b \cos x}$.Now,differentiating again with respect to $x$:$\frac{d^2 y}{d x^2} = \frac{d}{d x} (a+b \cos x)^{-1} = -(a+b \cos x)^{-2} \cdot (-b \sin x) = \frac{b \sin x}{(a+b \cos x)^2}$.At $x = \frac{\pi}{2}$:$\left.\frac{d^2 y}{d x^2}ight|_{x=\frac{\pi}{2}} = \frac{b \sin(\frac{\pi}{2})}{(a+b \cos(\frac{\pi}{2}))^2} = \frac{b(1)}{(a+0)^2} = \frac{b}{a^2}$.

If $y=\frac{2}{\sqrt{a^2-b^2}} \tan ^{-1}\left[\sqrt{\frac{a-b}{a+b}} \tan \frac{x}{2}\right]$,then $\left.\frac{d^2 y}{d x^2}\right|_{x=\frac{\pi}{2}}=$

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