$\frac{d}{d x}\left[\tan ^{-1}\left(\frac{a-x}{1+a x}\right)\right]$ is equal to
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$\frac{d}{d x}\left[\tan ^{-1}\left(\frac{a-x}{1+a x}\right)\right]$ is equal to

(A) $-\frac{1}{1+x^2}$

(B) $\frac{1}{1+a^2}-\frac{1}{1+x^2}$

(C) $\frac{1}{1+\left(\frac{a-x}{1+a x}\right)^2}$

(D) $\frac{-1}{\sqrt{1-\left(\frac{a-x}{1+a x}\right)^2}}$

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Best answer

SOLUTION —

$\text { } \begin{aligned}\frac{d}{d x}\left[\tan ^{-1}\left(\frac{a-x}{1+a x}\right)\right] & =\frac{d}{d x}\left[\tan ^{-1} a-\tan ^{-1} x\right] \\& =-\frac{1}{1+x^2}\end{aligned}$

So, The correct option will be (A).

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