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 - Page 136: 68

Answer

$f(x)=\dfrac{6-4x}{(x^2-3x)^3},f'(4)=-\dfrac{5}{32}.$

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.
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