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: 25


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

Work Step by Step

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