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: 62

Answer

$y'=\dfrac{\cos{\sqrt[3]{x}}}{3\sqrt[3]{x^2}}+\dfrac{\cos{x}}{3\sqrt[3]{sin^2{x}}}.$

Work Step by Step

$y=f(x)+g(x)\rightarrow f(x)=\sin{\sqrt[3]{x}}$; $g(x)=\sqrt[3]{\sin{x}}$ $f'(x)$ is found using the Chain Rule: $u=\sqrt[3]{x}$; $\dfrac{du}{dx}=\dfrac{1}{3\sqrt[3]{x^2}}$ $f(u)=\sin{u};\dfrac{d}{du}f(u)=\cos{u}$ $\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}=(\dfrac{1}{3\sqrt[3]{x^2}})(\cos{\sqrt[3]{x}})=\dfrac{\cos{\sqrt[3]{x}}}{3\sqrt[3]{x^2}}.$ $g'(x)$ is found using the Chain Rule: $u=\sin{x}$; $\dfrac{du}{dx}=\cos{x}$ $g(u)=u^{\frac{1}{3}};\dfrac{d}{du}g(u)=\dfrac{1}{3\sqrt[3]{u^2}}$ $\dfrac{d}{dx}g(x)=\dfrac{d}{du}g(u)\times\dfrac{du}{dx}=(\cos{x})(\dfrac{1}{3\sqrt[3]{sin^2{x}}})=\dfrac{\cos{x}}{3\sqrt[3]{sin^2{x}}}.$ $y'=f'(x)+g'(x)=\dfrac{\cos{\sqrt[3]{x}}}{3\sqrt[3]{x^2}}+\dfrac{\cos{x}}{3\sqrt[3]{sin^2{x}}}.$
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