Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.3 The Chain Rule - 4.3 Exercises - Page 225: 39


$\displaystyle \frac{dy}{dx}=\frac{-18x^{2}+2x+1}{(2x-1)^{6}}$

Work Step by Step

$y=\displaystyle \frac{3x^{2}-x}{(2x-1)^{5}}= \frac{u(x)}{v(x)}$ Use the quotient rule $\displaystyle \frac{dy}{dx}=\frac{v(t)\cdot u^{\prime}(t)-u(t)\cdot v^{\prime}(t)}{[v(t)]^{2}}$ For $v^{\prime}(x)$, we need the chain rule: $v^{\prime}(x)=[(2x-1)^{5}]^{\prime} =5(2x-1)^{4}\cdot(2x-1)^{\prime}=5(2x-1)^{4}\cdot 2]$ $\displaystyle \frac{dy}{dx}=\frac{(2x-1)^{5}(6x-1)-(3x^{2}-x)[5(2x-1)^{4}\cdot 2]}{[(2x-1)^{5}]^{2}}$ $=\displaystyle \frac{(2x-1)^{5}(6x-1)-10(3x^{2}-x)(2x-1)^{4}}{(2x-1)^{10}}$ ... factor out $(2x-1)^{4}$ in the numerator... $=\displaystyle \frac{(2x-1)^{4}[(2x-1)(6x-1)-10(3x^{2}-x)]}{(2x-1)^{10}}$ ... reduce the ratio with $(2x-1)^{4}$, simplify $=\displaystyle \frac{12x^{2}-2x-6x+1-30x^{2}+10x}{(2x-1)^{6}}$ $\displaystyle \frac{dy}{dx}=\frac{-18x^{2}+2x+1}{(2x-1)^{6}}$
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.