Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.5 Higher Derivatives - Exercises - Page 135: 7

Answer

$y' = 16x^{-1/5} -4x^{-1/3} $ $y'' = \dfrac{4}{3}x^{-4/3}-\dfrac{16}{5}x^{-6/5}$ $y''' = \dfrac{96}{25}x^{-11/5}-\dfrac{16}{9}x^{-7/3}$

Work Step by Step

$y' = 16x^{-1/5} -4x^{-1/3} $ $y'' = \dfrac{4}{3}x^{-4/3}-\dfrac{16}{5}x^{-6/5}$ $y''' = \dfrac{96}{25}x^{-11/5}-\dfrac{16}{9}x^{-7/3}$
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