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: 25

Answer

$\frac{dy}{dx}=\frac{-y^2-2xy}{x^2+2xy}$ Derivative at $(-1,1): -1$

Work Step by Step

$(x+y)^3=x^3+y^3$ $x^3+3x^2y+3xy^2+y^3=x^3+y^3$ $3x^2y+3xy^2=0$ $\frac{d}{dx}[3x^2y+3xy^2]=\frac{d}{dx}[0]$ $3(\frac{d}{dx}[x^2y+xy^2])=0$ $\frac{d}{dx}[x^2y+xy^2]=\frac{0}{3}$ $\frac{d}{dx}[x^2y+xy^2]=0$ $2xy+x^2\frac{dy}{dx}+y^2+2xy\frac{dy}{dx}=0$ $2xy+y^2+(x^2+2xy)\frac{dy}{dx}=0$ $(x^2+2xy)\frac{dy}{dx}=-y^2-2xy$ $\frac{dy}{dx}=\frac{-y^2-2xy}{x^2+2xy}$ $(-1,1)$ $\frac{-y^2-2xy}{x^2+2xy}=\frac{-(1)^2-2(-1)(1)}{(-1)^2+2(-1)(1)}=\frac{1}{-1}=-1$
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