Differentiate the following functions with respect to x :
Question:

Differentiate the following functions with respect to $x$ :

$\tan ^{-1}\left(\frac{a+b \tan x}{b-a \tan x}\right)$

Solution:

$y=\tan ^{-1}\left(\frac{a+b \tan x}{b-a \tan x}\right)$

Dividing numerator and denominator by $b$

$y=\tan ^{-1}\left(\frac{\frac{\mathrm{a}}{\mathrm{b}}+\tan \mathrm{x}}{1-\frac{\mathrm{a}}{\mathrm{b}} \tan \mathrm{x}}\right)$

$y=\tan ^{-1}\left(\frac{\tan \left(\tan ^{-1} \frac{\mathrm{a}}{\mathrm{b}}\right)+\tan \mathrm{x}}{1-\tan \left(\tan ^{-1} \frac{\mathrm{a}}{\mathrm{b}}\right) \tan \mathrm{x}}\right)$

Using, $\tan (\mathrm{x}+\mathrm{y})=\left(\frac{\tan \mathrm{x}+\tan \mathrm{y}}{1-\tan \mathrm{x} \tan \mathrm{y}}\right)$

$y=\tan ^{-1}\left(\tan \left(\tan ^{-1} \frac{a}{b}+x\right)\right)$

$y=\tan ^{-1} \frac{a}{b}+x$

Differentiating w.r.t $x$ we get

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\tan ^{-1} \frac{\mathrm{a}}{\mathrm{b}}+\mathrm{x}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}}=0+1$

$\frac{\mathrm{dy}}{\mathrm{dx}}=1$

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