Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 11 - Section 11.4 - The Chain Rule - Exercises - Page 831: 44

Answer

$\displaystyle \frac{|(x-3)^{1/3}|}{3(x-3)}$

Work Step by Step

Apply the Generalized Rule for absolute values, $\displaystyle \frac{d}{dx}|w|=\frac{|w|}{w}\frac{dw}{dx}=\frac{|(x-3)^{1/3}|}{(x-3)^{1/3}}\cdot\frac{dw}{dx}$ $w(x)=(x-3)^{1/3}$ $w$ is a composite function $u(v(x))$ where $u(x)=x^{1/3},\quad v(x)=x-3$ $\displaystyle \frac{du}{dv}=\frac{1}{3}v^{-2/3},\qquad \frac{dv}{dx}=1$ $\displaystyle \frac{dw}{dx}=\frac{du}{dv}\frac{dv}{dx}=\frac{1}{3}v^{-2/3}\cdot 1$ $=\displaystyle \frac{1}{3}(x-3)^{-2/3}$ $\displaystyle \frac{d}{dx}|w|=\frac{|(x-3)^{1/3}|}{(x-3)^{1/3}}\cdot\frac{1}{3}(x-3)^{-2/3}$ $=\displaystyle \frac{|(x-3)^{1/3}|}{3(x-3)^{1/3+2/3}}$ $=\displaystyle \frac{|(x-3)^{1/3}|}{3(x-3)}$
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