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

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

therefore,$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\sin \left( {{x^{1/2}}} \right) + \frac{d}{{dx}}{\left[ {\cos \left( {{x^{1/2}}} \right)} \right]^2}$

$= \cos {x^{1/2}} \cdot \frac{d}{{dx}}{x^{1/2}} + 2\cos \left( {{x^{1/2}}} \right)\frac{d}{{dx}}\left[ {\cos \left( {{x^{1/2}}} \right)} \right]$

$= \cos \left( {{x^{1/2}}} \right)\frac{1}{2}{x^{ - 1/2}} + 2 \cdot \cos \left( {{x^{1/2}}} \right) \cdot \left[ { - \sin \left( {{x^{1/2}}} \right) \cdot \frac{d}{{dx}}{x^{1/2}}} \right]$

$= \cos \sqrt x \cdot \frac{1}{{2\sqrt x }}\left[ { - 2\cos \left( {{x^{1/2}}} \right)} \right] \cdot \sin {x^{1/2}} \cdot \frac{1}{{2\sqrt x }}$

$= \frac{1}{{2\sqrt x }}[\cos (\sqrt x ) - \sin (2\sqrt x )]$

$\Rightarrow \frac{dy}{dx} = \frac{1}{{2\sqrt x }}[\cos (\sqrt x ) - \sin (2\sqrt x )]$