Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Review - Exercises - Page 267: 2



Work Step by Step

$y=\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt[5]{x^{3}}}$ Rewrite the denominators by expressing them as powers with rational exponents: $y=\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt[5]{x^{3}}}=\dfrac{1}{x^{1/2}}-\dfrac{1}{x^{3/5}}=...$ Now, change the sign of the exponents of the denominators by taking them to the numerator: $...=x^{-1/2}-x^{-3/5}$ Now, evaluate the derivate using the power rule: $y'=-\dfrac{1}{2}x^{-1/2-1}-\Big(-\dfrac{3}{5}\Big)x^{-3/5-1}=...$ $...=-\dfrac{1}{2}x^{-3/2}+\dfrac{3}{5}x^{-8/5}=...$ Give the answer without negative exponents by taking the powers to the denominator and changing the sign of the exponents: $...=-\dfrac{1}{2x^{3/2}}+\dfrac{3}{5x^{8/5}}=-\dfrac{1}{2\sqrt{x^{3}}}+\dfrac{3}{5\sqrt[5]{x^{8}}}$
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