Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.1 Derivative of f(x)=bx and the Number e - Exercises - Page 327: 39


$$ f'(x)= \frac{2 e^{x}-e^{x+1}-2 x e^{x}-1}{\left(2 e^{x}-1\right)^{2}}.$$

Work Step by Step

Recall that $(e^x)'=e^x$ Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Since we have $$ f(x)= \frac{e^{x+1}+x}{2e^x-1}$$ then the derivative $ f'(x)$, using the chain and quotient rules, is given by \begin{align*}f'(x)&=\frac{(2e^x-1)(e^{x+1}+x)'-(e^{x+1}+x)(2e^x-1)'}{(2e^x-1)^2} \\ &=\frac{(2e^x-1)(e^{x+1}+1)-(e^{x+1}+x)(2e^x)}{(2e^x-1)^2} \\ &= \frac{2 e^{x}-e^{x+1}-2 x e^{x}-1}{\left(2 e^{x}-1\right)^{2}}. \end{align*}
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