Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.5 Exercises: 29

Answer

$\dfrac{dy}{dx}=-\dfrac{2xy}{x^2+4}.$ At $(2, 1)\rightarrow\dfrac{dy}{dx}=-\dfrac{2(2)(1)}{2^2+4}=-\dfrac{1}{2}$

Work Step by Step

$\dfrac{d}{dx}((x^2+4)y)=\dfrac{d}{dx}(8)\rightarrow$ $y(\dfrac{d}{dx}(x^2+4))+(x^2+4)(\dfrac{d}{dx}(y))=0\rightarrow$ $2xy+\dfrac{dy}{dx}(x^2+4)=0\rightarrow$ $\dfrac{dy}{dx}=-\dfrac{2xy}{x^2+4}.$ At $(2, 1)\rightarrow\dfrac{dy}{dx}=-\dfrac{2(2)(1)}{2^2+4}=-\dfrac{1}{2}$
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