Question
Differentiate the following functions with respect to x:
$\log(\tan^{-1}\text{x})$

Answer

Consider $\text{y}=\log\big(\tan^{-1}\text{x}\big)$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\big(\tan^{-1}\text{x}\big)$
$=\frac{1}{\tan^{-1}\text{x}}\times\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{x}\big)$
[Using chain rule]
$=\frac{1}{\big(1+\text{x}^2\big)\tan^{-1}\text{x}}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}\big(\log\tan^{-1}\text{x}\big)=\frac{1}{\big(1+\text{x}^2\big)\tan^{-1}\text{x}}$

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