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: 12

Answer

$g'(x)=-\dfrac{3x}{\sqrt{4-3x^2}}$

Work Step by Step

$u=4-3x^2$; $\dfrac{du}{dx}=-6x$ $g(u)=\sqrt{u};\dfrac{d}{du}g(u)=\dfrac{1}{2\sqrt{u}}.$ $\dfrac{d}{dx}g(x)=\dfrac{d}{du}g(u)\times\dfrac{du}{dx}=(-6x)(\dfrac{1}{2\sqrt{4-3x^2}})=-\dfrac{3x}{\sqrt{4-3x^2}}$
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