Question
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{a}+\text{x}}{1-\text{ax}}\Big)$

Answer

Let $\text{y}=\tan^{-1}\Big(\frac{\text{a}+\text{x}}{1-\text{ax}}\Big)$
$\text{y}=\tan^{-1}\text{a}+\tan^{-1}\text{x}$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{a})+\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{x})$
$=0+\frac{1}{1+\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{x}^2}$

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