Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.9 Derivatives of Logarithmic and Exponential Functions - 3.9 Exercises: 62

Answer

\[{y^,} = 2x\cos x - {x^2}\sin x\]

Work Step by Step

\[\begin{gathered} f\,\left( x \right) = {x^2}\cos x \hfill \\ \hfill \\ f\,\left( x \right) = y\,\,\,then\,\,y = {x^2}\cos x \hfill \\ \hfill \\ Take\,\,\ln \,\,to\,\,each\,\,side \hfill \\ \hfill \\ \ln y = \ln \,\left( {{x^2}\cos x} \right) \hfill \\ \hfill \\ use\,\,\log \,\,properties \hfill \\ \hfill \\ \ln y = \ln {x^2} + \ln \cos x \hfill \\ \hfill \\ Differentiate \hfill \\ \hfill \\ \frac{{{y^,}}}{y} = \frac{{2x}}{{{x^2}}} + \frac{{ - \sin x}}{{\cos x}} \hfill \\ \hfill \\ Simplify \hfill \\ \hfill \\ \frac{{{y^,}}}{y} = \frac{2}{x} - \tan x \hfill \\ \hfill \\ multiply\,\,both\,\,\,sides\,\,by\,\,y \hfill \\ \hfill \\ {y^,} = \frac{{2y}}{x} - y\tan x \hfill \\ \hfill \\ Where\,\,\,y = {x^2}\cos x,\,\,then \hfill \\ \hfill \\ {y^,} = 2x\cos x - {x^2}\sin x \hfill \\ \end{gathered} \]
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