Prove the following

Let

$y=\sin \sqrt{x}+\cos ^{2} \sqrt{x}$

Differentiating both sides w.r.t. $x$

$\frac{d y}{d x}=\frac{d}{d x}(\sin \sqrt{x})+\frac{d}{d x}\left(\cos ^{2} \sqrt{x}\right)$

$=\cos \sqrt{x} \cdot \frac{d}{d x}(\sqrt{x})+2 \cos \sqrt{x} \cdot \frac{d}{d x}(\cos \sqrt{x})$

$=\cos \sqrt{x} \cdot \frac{1}{2 \sqrt{x}}+2 \cos \sqrt{x}(-\sin \sqrt{x}) \cdot \frac{d}{d x} \sqrt{x}$

$=\frac{1}{2 \sqrt{x}} \cdot \cos \sqrt{x}-2 \cos \sqrt{x} \cdot \sin \sqrt{x} \cdot \frac{1}{2 \sqrt{x}}$

$=\frac{\cos \sqrt{x}}{2 \sqrt{x}}-\frac{\sin 2 \sqrt{x}}{2 \sqrt{x}}$

Thus, $\quad \frac{d y}{d x}=\frac{\cos \sqrt{x}}{2 \sqrt{x}}-\frac{\sin 2 \sqrt{x}}{2 \sqrt{x}}$.