Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises: 11

Answer

$4x^3$

Work Step by Step

We are trying to find $\frac{d}{dx}(x-1)(x^3+x^2+x+1)$. Use the Product Rule, $\frac{d}{dx}u(x)v(x)=u'(x)v(x)+u(x)v'(x)$. In this case, $u(x)=x-1$, so $u'(x)=1$. Also, $v(x)=x^3+x^2+x+1$, so $v'(x)=3x^2+2x+1$. Plugging these in, we get: $\frac{d}{dw}(x-1)(x^3+x^2+x+1)$ $=1*(x^3+x^2+x+1)+(x-1)(3x^2+2x+1)$ $=(x^3+x^2+x+1)+(3x^3+2x^2+x)-(3x^2+2x+1)$ $=x^3+x^2+x+1+3x^3+2x^2+x-3x^2-2x-1$ $=4x^3$
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