Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises - Page 158: 17



Work Step by Step

$$f'(x)=((2x+3)^4(x^2+x+1)^5)'=((2x+3)^4)'\cdot(x^2+x+1)^5+(2x+3)^4((x^2+x+1)^5)'=4(2x+3)^3(2x+3)'(x^2+x+1)^5+(2x+3)^45(x^2+x+3)^4(x^2+x+1)'= 4(2x+3)^32(x^2+x+1)^5+5(2x+3)^4(x^2+x+1)^4(2x+1)=(2x+3)^3(x^2+x+1)^4(8(x^2+x+1)+5(2x+3)(2x+1))=(2x+3)^3(x^2+x+1)^4(8x^2+8x+8+5(4x^2+8x+3))=(2x+3)^3(x^2+x+1)^4(28x^2+48x+23)$$
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.