University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 158: 35

Answer

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

Work Step by Step

$$y=xe^{-x}+e^{x^3}$$ The derivative of function $y$ is: $$y'=(xe^{-x})'+(e^{x^3})'=(x)'e^{-x}+x(e^{-x})'+\Big(e^{x^3}(x^3)'\Big)$$ $$y'=e^{-x}+xe^{-x}(-x)'+\Big(e^{x^3}(3x^2)\Big)$$ $$y'=e^{-x}+xe^{-x}(-1)+3x^2e^{x^3}$$ $$y'=e^{-x}-xe^{-x}+3x^2e^{x^3}$$ $$y'=e^{-x}(1-x)+3x^2e^{x^3}$$
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