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: 38



Work Step by Step

$$y=(9x^2-6x+2)e^{x^3}$$ The derivative of function $y$ is: $$y'=(9x^2-6x+2)'e^{x^3}+(9x^2-6x+2)(e^{x^3})'$$ $$y'=(18x-6)e^{x^3}+(9x^2-6x+2)e^{x^3}(x^3)'$$ $$y'=(18x-6)e^{x^3}+3x^2(9x^2-6x+2)e^{x^3}$$ $$y'=(18x-6)e^{x^3}+(27x^4-18x^3+6x^2)e^{x^3}$$ $$y'=e^{x^3}(27x^4-18x^3+6x^2+18x-6)$$ $$y'=3e^{x^3}(9x^4-6x^3+2x^2+6x-2)$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.