frac{d}{dx} tan^{-1} left( frac{1-x}{1+x} rig | vedclass

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(A) Let $y = 	an^{-1} \left( \frac{1-x}{1+x} ight)$.We know that $	an^{-1} \left( \frac{a-b}{1+ab} ight) = 	an^{-1} a - 	an^{-1} b$.Here,$a =

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $y = 	an^{-1} \left( \frac{1-x}{1+x} ight)$.We know that $	an^{-1} \left( \frac{a-b}{1+ab} ight) = 	an^{-1} a - 	an^{-1} b$.Here,$a = 1$ and $b = x$,so $y = 	an^{-1}(1) - 	an^{-1}(x)$.Since $	an^{-1}(1) = \frac{\pi}{4}$,we have $y = \frac{\pi}{4} - 	an^{-1}(x)$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} ight) - \frac{d}{dx} \left( 	an^{-1} x ight)$.$\frac{dy}{dx} = 0 - \frac{1}{1+x^2} = \frac{-1}{1+x^2}$.Thus,the correct option is $A$.

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

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