Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises: 46

Answer

$y'= 6x-2-2e^{-2x}$

Work Step by Step

$y = 3x^2-2x+e^{-2x}$ Using Power Rule and $\frac{d}{dx}e^{nx}$ = $ne^{nx}$: $y'=3(2)x^{(2-1)} - 2(1)x^{(1-1)}+(-2)e^{-2x} = 6x-2-2e^{-2x}$
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.