University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.3 - Differentiation Rules - Exercises: 30

Answer

$\displaystyle y' = \frac{e^{x}(x+3) - x^{2}(1+2e^{x})}{(2e^{x}-x)^{2}}$

Work Step by Step

$y = \frac{x^{2}+3e^{x}}{2e^{x}-x}$ $y' = \frac{(2e^{x}-x)(2x +3e^{x}) - (x^{2}+3e^{x})(2e^{x}-1)}{(2e^{x}-x)^{2}}$ $y' = \frac{xe^{x}-x^{2}-2x^{2}e^{x}+3e^{x}}{(2e^{x}-x)^{2}}$ $y' = \frac{e^{x}(x+3) - x^{2}(1+2e^{x})}{(2e^{x}-x)^{2}}$
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