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: 22

Answer

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

Work Step by Step

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