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: 17


$h^{\prime}(x)=-\displaystyle \frac{6x}{(x^{2}+1)^{4}}$

Work Step by Step

We can rewrite $h(x)=(x^{2}+1)^{-3}$ Let $u(x)=x^{-3},\qquad v(x)=x^{2}+1$ $\displaystyle \frac{du}{dx}=-3x^{-4}, \quad \frac{dv}{dx}=2x$ Then,$\quad y=h(x)=u(v(x))$ and $f^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$ $\displaystyle \frac{du}{dv}=-3v^{-4}=-3(x^{2}+1)^{-4}$ $h^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}=-3(x^{2}+1)^{-4}\cdot(2x)$ $h^{\prime}(x)=-6x(x^{2}+1)^{-4}$ $h^{\prime}(x)=-\displaystyle \frac{6x}{(x^{2}+1)^{4}}$
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