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: 19

Answer

$f'(t)=-\dfrac{2}{(t-3)^3}.$

Work Step by Step

$f(t)=\dfrac{1}{(t-3)^2}=(t-3)^{-2}.$ Using the Chain Rule: $u=(t-3)$; $\dfrac{du}{dt}=1.$ $f(u)=u^{-2}$ $\dfrac{d}{du}f(u)=-\dfrac{2}{u^3}=-\dfrac{2}{(t-3)^3}.$ $\dfrac{d}{dt}f(t)=\dfrac{d}{du}f(u)\times\dfrac{du}{dt}=-\dfrac{2}{(t-3)^3}.$
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