frac{d}{dx} left{ sin^2 left( cot^{-1} sqrt{f | vedclass

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

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$\frac{- 1}{2}$

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(B) Let $y = \sin^2 \left( \cot^{-1} \sqrt{\frac{1 + x}{1 - x}} ight)$.Substitute $x = \cos 2	heta$,so $	heta = \frac{1}{2} \cos^{-1} x$.Then $\sqrt{\frac{1 + x}{1 - x}} = \sqrt{\frac{1 + \cos 2	heta}{1 - \cos 2	heta}} = \sqrt{\frac{2 \cos^2 	heta}{2 \sin^2 	heta}} = \cot 	heta$.Thus,$y = \sin^2 \left( \cot^{-1} (\cot 	heta) ight) = \sin^2 	heta$.Using the identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we get $y = \frac{1 - x}{2} = \frac{1}{2} - \frac{x}{2}$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{2} - \frac{x}{2} ight) = 0 - \frac{1}{2} = -\frac{1}{2}$.

$\frac{d}{dx} \left\{ \sin^2 \left( \cot^{-1} \sqrt{\frac{1 + x}{1 - x}} \right) \right\} =$

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