Question:
Differentiate the following functions with respect to $x$ :
$\tan ^{-1}\left(\frac{a+x}{1-a x}\right)$
Solution:
$y=\tan ^{-1}\left(\frac{a+x}{1-a x}\right)$
Using, $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$y=\tan ^{-1} x+\tan ^{-1} a$
Differentiating w.r.t $\mathrm{x}$ we get
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{a}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+\mathrm{x}^{2}}+0$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+\mathrm{x}^{2}}$
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