Answer
$(sin2^{x})+x.2^{x}(cos2^{x})ln2$
Work Step by Step
$\frac{d}{dx}g(x)=\frac{d}{dx}(x sin2^{x})$
$=\frac{d}{dx}(x)(sin2^{x})+x\frac{d}{dx}(sin2^{x})$
$=1.(sin2^{x})+x.(cos2^{x})(2^{x})ln2$
$=(sin2^{x})+x.2^{x}(cos2^{x})ln2$
Calculus 8th Edition
by Stewart, James
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 - Page 436: 18
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