Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.4 The Product And Quotient Rules - Exercises Set 2.4 - Page 147: 5



Work Step by Step

According to the product rule, if f(x)=h(x)g(x), then f'(x)=h(x)g'(x)+g(x)h'(x). If we set h(x)=3$x^{2}$+6, and g(x)=2x-$\frac{1}{4}$, then using the product rule, f'(x)=(3$x^{2}$+6)$\frac{d}{dx}$[2x-$\frac{1}{4}$]+(2x-$\frac{1}{4}$)$\frac{d}{dx}$[3$x^{2}$+6] $\frac{d}{dx}$[2x-$\frac{1}{4}$]=2, using the power rule $\frac{d}{dx}$[3$x^{2}$+6]=6x, using the power rule Therefore, f'(x)=(3$x^{2}$+6)(2)+(6x)(2x-$\frac{1}{4}$) = 6$x^{2}$+12+12$x^{2}$-$\frac{3x}{2}$ =18$x^{2}$-$\frac{3x}{2}$+12
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