$\begin{align}  & y=\sec \sqrt{x} \\ & \text{Let }u=\sqrt{x}={{x}^{\tfrac{1}{2}}},\text{  }\frac{du}{dx}=\frac{1}{2}{{x}^{-\tfrac{1}{2}}} \\ & y=\sec u \\ & \frac{dy}{du}=\sec u\tan u \\ & \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}=\sec u\tan u\times \left( \frac{1}{2}{{x}^{-\tfrac{1}{2}}} \right) \\ & \frac{dy}{dx}=\frac{1}{2\sqrt{x}}\sec \sqrt{x}\tan \sqrt{x} \\\end{align}$