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