Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.4 Exercises - Page 136: 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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.