Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 17

Answer

$f'(x)=(2x-3)^{3}(x^{2}+x+1)^{4}[5(2x-3)(2x+1)+8(x^{2}+x+1)]$

Work Step by Step

$f(x)=(2x-3)^{4}(x^{2}+x+1)^{5}$ Differentiate using the product rule: $f'(x)=[(2x-3)^{4}][(x^{2}+x+1)^{5}]'+[(x^{2}+x+1)^{5}][(2x-3)^{4}]'=...$ Use the chain rule to find $[(x^{2}+x+1)^{5}]'$ and $[(2x-3)^{4}]'$: $...=[(2x-3)^{4}][5(x^{2}+x+1)^{4}(x^{2}+x+1)']+[(x^{2}+x+1)^{5}][4(2x-3)^{3}(2x-3)']=...$ $...=5(2x-3)^{4}(x^{2}+x+1)^{4}(2x+1)+4(x^{2}+x+1)^{5}(2x-3)^{3}(2)=...$ $...=5(2x-3)^{4}(x^{2}+x+1)^{4}(2x+1)+8(x^{2}+x+1)^{5}(2x-3)^{3}$ Take out common factor $(2x-3)^{3}$ and $(x^{2}+x+1)^{4}$: (Optional) $f'(x)=(2x-3)^{3}(x^{2}+x+1)^{4}[5(2x-3)(2x+1)+8(x^{2}+x+1)]$
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.