If y = tan^{-1} left{ frac{sqrt{1 + x^2} - sq | vedclass

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(B) Let $x^2 = \cos 	heta$, so $	heta = \cos^{-1}(x^2)$.Then $y = 	an^{-1} \left\{ \frac{\sqrt{1 + \cos 	heta} - \sqrt{1 - \cos 	heta}}{\sqrt{1 +

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$\frac{x}{\sqrt{1 - x^4}}$

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(B) Let $x^2 = \cos 	heta$, so $	heta = \cos^{-1}(x^2)$.Then $y = 	an^{-1} \left\{ \frac{\sqrt{1 + \cos 	heta} - \sqrt{1 - \cos 	heta}}{\sqrt{1 + \cos 	heta} + \sqrt{1 - \cos 	heta}} ight\}$.Using half-angle formulas $\sqrt{1 + \cos 	heta} = \sqrt{2} \cos(	heta/2)$ and $\sqrt{1 - \cos 	heta} = \sqrt{2} \sin(	heta/2)$, we get:$y = 	an^{-1} \left\{ \frac{\sqrt{2} \cos(	heta/2) - \sqrt{2} \sin(	heta/2)}{\sqrt{2} \cos(	heta/2) + \sqrt{2} \sin(	heta/2)} ight\} = 	an^{-1} \left\{ \frac{1 - 	an(	heta/2)}{1 + 	an(	heta/2)} ight\}$.$y = 	an^{-1} \left\{ 	an(\frac{\pi}{4} - \frac{	heta}{2}) ight\} = \frac{\pi}{4} - \frac{	heta}{2} = \frac{\pi}{4} - \frac{1}{2} \cos^{-1}(x^2)$.Differentiating with respect to $x$:$\frac{dy}{dx} = 0 - \frac{1}{2} \left( -\frac{1}{\sqrt{1 - (x^2)^2}} \cdot 2x ight) = \frac{x}{\sqrt{1 - x^4}}$.

If $y = \tan^{-1} \left\{ \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} \right\}$, where $|x| < 1$, then $\frac{dy}{dx}$ is equal to

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