Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Review - Exercises: 24

Answer

$y'=\dfrac{1+\dfrac{1}{2x^{1/2}}}{3(x+x^{1/2})^{4/3}}=\dfrac{2x^{1/2}+1}{6x^{1/2}(x+x^{1/2})^{4/3}}$

Work Step by Step

$y=1/\sqrt[3]{x+\sqrt{x}}$ Start differentiating by using the quotient rule: $y'=\dfrac{(\sqrt[3]{x+\sqrt{x}})(1)'-(1)(\sqrt[3]{x+\sqrt{x}})'}{(\sqrt[3]{x+\sqrt{x}})^{2}}=...$ $...=\dfrac{(\sqrt[3]{x+\sqrt{x}})(0)-(1)(\sqrt[3]{x+\sqrt{x}})'}{(\sqrt[3]{x+\sqrt{x}})^{2}}=\dfrac{-(\sqrt[3]{x+\sqrt{x}})'}{(\sqrt[3]{x+\sqrt{x}})^{2}}=...$ Rewrite $\sqrt[3]{x+\sqrt{x}}$ using rational exponents: $...=\dfrac{-[(x+x^{1/2})^{1/3}]'}{[(x+x^{1/2})^{1/3}]^{2}}=\dfrac{-[(x+x^{1/2})^{1/3}]'}{(x+x^{1/2})^{2/3}}=...$ Use the chain rule to evaluate the derivative indicated in the numerator: $...=\dfrac{-\Big[\dfrac{1}{3}(x+x^{1/2})^{-2/3}(x+x^{1/2})'\Big]}{(x+x^{1/2})^{2/3}}=...$ Evaluate the last derivative and simplify: $...=-\dfrac{(x+x^{1/2})^{-2/3}\Big(1+\dfrac{1}{2}x^{-1/2}\Big)}{3(x+x^{1/2})^{2/3}}=\dfrac{1+\dfrac{1}{2x^{1/2}}}{3(x+x^{1/2})^{4/3}}=...$ $...=\dfrac{\dfrac{2x^{1/2}+1}{2x^{1/2}}}{3(x+x^{1/2})^{4/3}}=\dfrac{2x^{1/2}+1}{6x^{1/2}(x+x^{1/2})^{4/3}}$
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