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: 43

Answer

$ \displaystyle \frac{6x^{2}-4x -7}{(3x-1)^{2}}$

Work Step by Step

$ \displaystyle \frac{d}{dx}[\frac{f(x)}{g(x)}]= \displaystyle \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$ $f(x)=2x^{2}+4x+1,\quad f^{\prime}(x)=4x+4$ $g(x)=3x-1,\qquad g^{\prime}(x)=3$ $\displaystyle \frac{dy}{dx}=\frac{(4x+4)(3x-1)-(2x^{2}+4x+1)(3)}{(3x-1)^{2}}$ $=\displaystyle \frac{(12x^{2}-4x+12x-4)-(6x^{2}+12x+3)}{(3x-1)^{2}}$ $=\displaystyle \frac{6x^{2}-4x -7}{(3x-1)^{2}}$
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