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: 61

Answer

$y'=\dfrac{1}{2\sqrt{x}}+2x\cos{(2x)}^2.$

Work Step by Step

$y=f(x)+g(x)\rightarrow f(x)=\sqrt{x}$ ; $g(x)=\frac{1}{4}\sin{(2x)}^2$ $f'(x)=\dfrac{1}{2\sqrt{x}}.$ $g'(x)$ is found using the Chain Rule: $u=(2x)^2$; $\dfrac{du}{dx}=8x$ $g(u)=\frac{1}{4}\sin{u};\dfrac{d}{du}g(u)=\frac{1}{4}\cos{u}$ $\dfrac{d}{dx}g(x)=\dfrac{d}{du}g(u)\times\dfrac{du}{dx}=(8x)(\frac{1}{4}\cos{(2x)}^2)=2x\cos{(2x)}^2.$ $y'=f'(x)+g'(x)=\dfrac{1}{2\sqrt{x}}+2x\cos{(2x)}^2.$
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