Find \frac{d y}{d x} if y=\sec (\tan \sqrt{x}).

    \[y=\sec (\sqrt{\tan x})\]

applying chain nule

    \[\begin{aligned} \frac{d y}{d x} &=(\sec \sqrt{\tan n} \times \tan \sqrt{\tan n}) \times \frac{1}{2}(\tan n)^{1 / 2-1} \times \sec ^{2} n \\ &=\frac{\sec \sqrt{\tan n} \tan \sqrt{\tan n}}{2 \sqrt{\tan n}} \times \sec ^{2} n . \end{aligned}\]