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: 14



Work Step by Step

$u=(x^2-4x+2)$; $\dfrac{du}{dx}=(2x-4)$ $f(u)=\sqrt{u}; \dfrac{d}{du}f(u)=\dfrac{1}{2\sqrt{u}}$ $\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}$ $=(\dfrac{1}{2\sqrt{(x^2-4x+2)}})(2x-4)$ $=\dfrac{x-2}{\sqrt{x^2-4x+2}}$
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