Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 11 - Section 11.4 - The Chain Rule - Exercises: 12

Answer

$f^{\prime}(x)=3(x^{3}-x)^{2}\cdot(3x^{2}-1)$

Work Step by Step

Let $u(x)=x^{3},\qquad v(x)=x^{3}-x$ $\displaystyle \frac{du}{dx}=3x^{2}, \quad \frac{dv}{dx}=3x^{2}-1$ Then,$\quad y=f(x)=u(v(x))$ and $f^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$ $\displaystyle \frac{du}{dv}=3v^{2}=3(x^{3}-x)^{2}$ $f^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}=3(x^{3}-x)^{2}\cdot(3x^{2}-1)$ $f^{\prime}(x)=3(x^{3}-x)^{2}\cdot(3x^{2}-1)$
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