Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.3 Product and Quotient Rules - Exercises: 4

Answer

f'(x)=14$x^{6}$-24$x^{5}$+15$x^{4}$-24$x^{3}$

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)=(3$x^{4}$+2$x^{6}$)(x-2) by setting g(x)=3$x^{4}$+2$x^{6}$ and h(x)=x-2, and applying the product rule. f'(x)=$\frac{d}{dx}$[3$x^{4}$+2$x^{6}$](x-2)+(3$x^{4}$+2$x^{6}$)$\frac{d}{dx}$[x-2] $\frac{d}{dx}$[3$x^{4}$+2$x^{6}$]=12$x^{3}$+12$x^{5}$, using the power rule $\frac{d}{dx}$[x-2]=1, using the power rule Therefore, f'(x)=(12$x^{3}$+12$x^{5}$)(x-2)+(3$x^{4}$+2$x^{6}$)(1) =(12$x^{4}$+12$x^{6}$-24$x^{3}$-24$x^{5}$)+(3$x^{4}$+2$x^{6}$) (Simplify) =14$x^{6}$-24$x^{5}$+15$x^{4}$-24$x^{3}$ (Simplify)
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