Answer
$y'=-\dfrac{\sin{x}}{2\sqrt{\cos{x}}}-\dfrac{1}{x^2}.$
Derivative not defined at point $(\dfrac{\pi}{2}, \dfrac{2}{\pi}).$
Work Step by Step
$y=f(x)+g(x)\rightarrow g(x)=\sqrt{\cos{x}}$ ; $f(x)=\dfrac{1}{x}$
$f'(x)$ is found using the Power Rule:
$f'(x)=-\dfrac{1}{x^2}$
$g'(x)$ is found using the Chain Rule:
$u=\cos{x}$; $\dfrac{du}{dx}=-\sin{x}$
$g(u)=\sqrt{u};\dfrac{d}{du}g(u)=\dfrac{1}{2\sqrt{u}}$
$\dfrac{d}{dx}g(x)=\dfrac{d}{du}g(u)\times\dfrac{du}{dx}=-\dfrac{\sin{x}}{2\sqrt{\cos{x}}}.$
$y'=f'(x)+g'(x)=-\dfrac{\sin{x}}{2\sqrt{\cos{x}}}-\dfrac{1}{x^2}.$
To evaluate the derivative, plug in $x=\dfrac{\pi}{2}\rightarrow$ $\dfrac{\sin{\dfrac{\pi}{2}}}{2\sqrt{\cos{\dfrac{\pi}{2}}}}-\dfrac{1}{(\dfrac{\pi}{2})^2}\rightarrow$
Notice that we get division by $0$, hence the derivative is not defined at the point and so it doesn't exist.
A graphing utility was used to verify this result.
