Find $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}$, if $\displaystyle \mathrm{y}=(\cos x)^{x}+\cos ^{-1} \sqrt{x}$ is given.

Let $\displaystyle \mathrm{u}=(\cos x)^{x} \Rightarrow \frac{d \mathrm{u}}{d x}=(\cos x)^{x}(-x \tan x+\log (\cos x))$,\mathrm{v}=\cos ^{-1} \sqrt{x} \Rightarrow \frac{d \mathrm{v}}{d x}=\frac{-1}{2 \sqrt{x-x^{2}}}Since, $\displaystyle \mathrm{y}=\mathrm{u}+\mathrm{v} \Rightarrow \frac{d \mathrm{y}}{d x}=\frac{d \mathrm{u}}{d x}+\frac{d \mathrm{v}}{d x}=(\cos x)^{x}(-x \tan x+\log (\cos x))+\frac{-1}{2 \sqrt{x-x^{2}}}$