frac{d}{dx} left{ cos^{-1} left( frac{1 - x^2 | vedclass

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(D) Let $y = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} ight)$.Substitute $x = 	an 	heta$,where $	heta = 	an^{-1} x$.Then,$\frac{1 - x^2}{1 + x^2}

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $y = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} ight)$.Substitute $x = 	an 	heta$,where $	heta = 	an^{-1} x$.Then,$\frac{1 - x^2}{1 + x^2} = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} = \cos(2	heta)$.So,$y = \cos^{-1}(\cos(2	heta)) = 2	heta = 2 	an^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = 2 \cdot \frac{d}{dx}(	an^{-1} x) = 2 \cdot \frac{1}{1 + x^2} = \frac{2}{1 + x^2}$.

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

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