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 - Page 831: 18


$h^{\prime}=-\displaystyle \frac{4x+2}{(x^{2}+x+1)^{3}}$

Work Step by Step

We can rewrite $h(x)=(x^{2}+x+1)^{-2}$ (h is a composite function) Let $u(x)=x^{-2},\qquad v(x)=x^{2}+x+1$ Then,$\quad y=h(x)=u(v(x))$ and $h^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$ $\displaystyle \frac{du}{dx}=-2x^{-3}, \quad \frac{dv}{dx}=2x+1$ $\displaystyle \frac{du}{dv}=-2v^{-3}=-2(x^{2}+x+1)^{-3}$ $h^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}=-2(x^{2}+x+1)^{-3}\cdot(2x+1)$ $h^{\prime}(x)=-2(2x+1)(x^{2}+x+1)^{-3}$ $h^{\prime}(x)=-\displaystyle \frac{4x+2}{(x^{2}+x+1)^{3}}$
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