Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.4 Exercises: 68



Work Step by Step

$f(x)=(x^2-3x)^{-2}$ $u=x^2-3x$; $\dfrac{du}{dx}=2x-3$ $f(u)=u^{-2};\dfrac{d}{du}f(u)=-\dfrac{2}{u^3}$ $\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}=\dfrac{6-4x}{(x^2-3x)^3}.$ $f'(4)=\dfrac{6-4(4)}{(4^2-3(4))^3}=-\dfrac{5}{32}.$ A graphing utility was used to verify this result.
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.