## Calculus 8th Edition

Published by Cengage

# Chapter 6 - Inverse Functions - 6.4 Derivatives of Logarithmic Functions - 6.4 Exercises: 18

#### Answer

$(sin2^{x})+x.2^{x}(cos2^{x})ln2$

#### Work Step by Step

$\frac{d}{dx}g(x)=\frac{d}{dx}(x sin2^{x})$ $=\frac{d}{dx}(x)(sin2^{x})+x\frac{d}{dx}(sin2^{x})$ $=1.(sin2^{x})+x.(cos2^{x})(2^{x})ln2$ $=(sin2^{x})+x.2^{x}(cos2^{x})ln2$

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