Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.5 Exercises: 46

Answer

$g'(α)=(5^{-\frac{α}{2}})((-\frac{1}{2}\log 5)(sin(2α))+2cos(2α))$

Work Step by Step

$g(α)= 5^{-\frac{α}{2}}sin2α$ $u=5^{-\frac{α}{2}}$ $u’=(-\frac{1}{2}\ln 5)5^{-\frac{α}{2}}$ $v=sin2α$ $v’=2cos2α$ $g’(α)=u’v+uv’$ $=((-\frac{1}{2}\ln 5)5^{-\frac{α}{2}})(sin(2α))+(5^{-\frac{α}{2}})(2cos(2α))$ $=(5^{-\frac{α}{2}})((-\frac{1}{2}\log 5)(sin(2α))+2cos(2α))$
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