Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 386: 15


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

Work Step by Step

Recall that $(e^x)'=e^x$ Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Since $ f(x)= \frac{e^{-x}}{x}$, then the derivative, by using the quotient rule, is given by $$ f'(x)=\frac{e^{-x}-x e^{-x} (-x)'}{x^2}=\frac{e^{-x}+x e^{-x} }{x^2}=\frac{(1+x )e^{-x} }{x^2}.$$
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