Question
Diffrentiate the following w.r.t.x
$\sqrt{\tan \sqrt{x}}$
$\sqrt{\tan \sqrt{x}}$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(\sqrt{\tan \sqrt{x}}) \\ & =\frac{1}{2 \sqrt{\tan \sqrt{x}}} \cdot \frac{d}{d x}(\tan \sqrt{x}) \\ & =\frac{1}{2 \sqrt{\tan \sqrt{x}}} \times \sec ^2 \sqrt{x} \cdot \frac{d}{d x}(\sqrt{x}) \\ & =\frac{1}{2 \sqrt{\tan \sqrt{x}}} \times \sec ^2 \sqrt{x} \times \frac{1}{2 \sqrt{x}} \\ & =\frac{\sec ^2 \sqrt{x}}{4 \sqrt{x} \sqrt{\tan \sqrt{x}}} .\end{aligned}$
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