Calculus (3rd Edition)

Published by W. H. Freeman

Chapter 3 - Differentiation - 3.3 Product and Quotient Rules - Exercises - Page 121: 2

Answer

f'(x)=18$x^{2}$-20x-9

Work Step by Step

The product rule states that if f(x)=h(x)g(x), then f'(x)=g'(x)h(x)+g(x)h'(x). We can find the derivative of the function f(x)=(3x-5)($2x^{2}-3$) by setting g(x)=3x-5 and h(x)=$2x^{2}-3$, and applying the product rule. f'(x)=$\frac{d}{dx}$[3x-5]($2x^{2}-3$)+(3x-5)$\frac{d}{dx}$[$2x^{2}-3$] $\frac{d}{dx}$[3x-5]=3, using the power rule $\frac{d}{dx}$[$2x^{2}-3$]=4x, using the power rule Therefore, f'(x)=(3)($2x^{2}-3$)+(3x-5)(4x) =(6$x^{2}$-9)+(12$x^{2}$-20x) (Simplify) =18$x^{2}$-20x-9 (Simplify)

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