Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.4 Derivatives of Logarithmic Functions - 6.4 Exercises: 11

Answer

$\ln t(\ln t\cos t+\frac{2\sin t}{t})$

Work Step by Step

$\frac{d}{dt}(\ln t)^2\sin t$ $=(\ln t)^2\frac{d}{dt}\sin t+\sin t\frac{d}{dt}(\ln t)^2$ $=(\ln t)^2\cos t+\sin t*2\ln t\frac{d}{dt}\ln t$ $=(\ln t)^2\cos t+2\sin t\ln t*\frac{1}{t}$ $=\ln t(\ln t\cos t+\frac{2\sin t}{t})$
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