Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 47


$$ y'=(2^x\ln 2+3^{-x}\ln 3)x^{-1} - (2^x-3^{-x} )x^{-2}.$$

Work Step by Step

Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(a^x)'=a^x\ln{a}$ Since $ y=\frac{2^x-3^{-x}}{x}$, it is easier to simplify $ y $ as follows $$ y=(2^x-3^{-x})x^{-1} $$ then, by the product rule, we have $$ y'=(2^x\ln 2+3^{-x}\ln 3)x^{-1} - (2^x-3^{-x} )x^{-2}.$$
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