# Chapter 2 - Differentiation - 2.4 Exercises: 72

$y'=-\dfrac{\sin{x}}{2\sqrt{\cos{x}}}-\dfrac{1}{x^2}.$ 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.

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