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: 6



Work Step by Step

f(x)=(2-x-3$x^{3}$)(7+$x^{5})$ 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)=2-x-3$x^{3}$ and g(x)=7+$x^{5}$, then f'(x)=(2-x-3$x^{3}$)$\frac{d}{dx}$[7+$x^{5}$]+(7+$x^{5}$)$\frac{d}{dx}$[2-x-3$x^{3}$]. $\frac{d}{dx}$[2-x-3$x^{3}]$=-9$x^{2}$-1, using the power rule $\frac{d}{dx}$[7+$x^{5}]$=5$x^{4}$, using the power rule Therefore, f'(x)=(2-x-3$x^{3}$)(5$x^{4})$+(7+$x^{5}$)(-9$x^{2}$-1) =10$x^{4}$-5$x^{5}$-15$x^{7}$+(-63$x^{2}$-7-9$x^{7}$-$x^{5})$ =-24$x^{7}$-6$x^{5}$+10$x^{4}$-63$x^{2}$-7
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