Question
Diffrentiate the following w. r. t. x.
$\tan ^{-1}(\log x)$
$\tan ^{-1}(\log x)$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\tan ^{-1}(\log x)\right] \\ & =\frac{1}{1+(\log x)^2} \cdot \frac{d}{d x}(\log x) \\ & =\frac{1}{1+(\log x)^2} \times \frac{1}{x} \\ & =\frac{1}{x\left[1+(\log x)^2\right]} .\end{aligned}$
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