Finite Math and Applied Calculus (6th Edition)

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

Chapter 11 - Section 11.3 - The Product and Quotient Rules - Exercises - Page 818: 20

Answer

$\displaystyle \frac{dy}{dx}=\frac{6}{(2-3x )^{2}}$

Work Step by Step

$f(x)=3x,\displaystyle \ \ \quad g(x)=2-3x,\quad y=\frac{f(x)}{g(x)}$ $f^{\prime}(x)=3\qquad g^{\prime}(x)=-3$ $\displaystyle \frac{dy}{dx}=\frac{d}{dx}[\frac{f(x)}{g(x)}]$= ... quotient rule ... $= \displaystyle \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$ $=\displaystyle \frac{3(2-3x ) - 3x(-3)}{(2-3x )^{2}}$ $=\displaystyle \frac{6-9x+9x}{(2-3x )^{2}}$ $= \displaystyle \frac{6}{(2-3x )^{2}}$
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