If y = cot^{-1} left( frac{1 + x}{1 - x} righ | vedclass

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(B) Given $y = \cot^{-1} \left( \frac{1 + x}{1 - x} ight)$.We know that $\frac{1 + x}{1 - x} = 	an \left( \frac{\pi}{4} + 	an^{-1} x ight)$.Thus

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$-\frac{1}{1 + x^2}$

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(B) Given $y = \cot^{-1} \left( \frac{1 + x}{1 - x} ight)$.We know that $\frac{1 + x}{1 - x} = 	an \left( \frac{\pi}{4} + 	an^{-1} x ight)$.Thus,$y = \cot^{-1} \left( 	an \left( \frac{\pi}{4} + 	an^{-1} x ight) ight)$.Using $\cot^{-1} 	heta = \frac{\pi}{2} - 	an^{-1} 	heta$,we have $y = \frac{\pi}{2} - 	an^{-1} \left( 	an \left( \frac{\pi}{4} + 	an^{-1} x ight) ight)$.$y = \frac{\pi}{2} - \left( \frac{\pi}{4} + 	an^{-1} x ight) = \frac{\pi}{4} - 	an^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} - 	an^{-1} x ight) = 0 - \frac{1}{1 + x^2} = -\frac{1}{1 + x^2}$.

If $y = \cot^{-1} \left( \frac{1 + x}{1 - x} \right)$,then $\frac{dy}{dx} = $

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