frac{d}{dx} left[ tan^{-1} left( frac{a - x}{ | vedclass

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Question

(A) Let $y = 	an^{-1} \left( \frac{a - x}{1 + ax} ight)$.Using the trigonometric identity $	an^{-1} \left( \frac{A - B}{1 + AB} ight) = 	an^{-1

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

Vedclass · Answer

(A) Let $y = 	an^{-1} \left( \frac{a - x}{1 + ax} ight)$.Using the trigonometric identity $	an^{-1} \left( \frac{A - B}{1 + AB} ight) = 	an^{-1} A - 	an^{-1} B$,we can rewrite the expression as:$y = 	an^{-1} a - 	an^{-1} x$.Differentiating both sides with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} (	an^{-1} a) - \frac{d}{dx} (	an^{-1} x)$.Since $	an^{-1} a$ is a constant,its derivative is $0$.Therefore,$\frac{dy}{dx} = 0 - \frac{1}{1 + x^2} = -\frac{1}{1 + x^2}$.

$\frac{d}{dx} \left[ \tan^{-1} \left( \frac{a - x}{1 + ax} \right) \right] = $

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