Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.8 Implicit Differentiation - Exercises - Page 152: 36


$$y =3x-2$$

Work Step by Step

Given $$\frac{x}{x+1}+\frac{x}{y+1}=1, \quad(1,1)$$ Differentiate both sides \begin{align*} \frac{\frac{d}{dx}\left(x\right)\left(x+1\right)-\frac{d}{dx}\left(x+1\right)x}{\left(x+1\right)^2}+ \frac{(1+y)-xy'}{(1+y)^2} &=0 \\ \frac{1}{\left(x+1\right)^2}+\frac{1}{1+y} - \frac{x}{(1+y)^2} y' &=0\\ y'&= \frac{(1+y)^2}{x\left(x+1\right)^2}+\frac{1+y}{x} \\ \end{align*} Then $$m = 3 $$ Hence, the tangent line is given by \begin{align*} \frac{y-y_1}{x-x_1}&=m\\ \frac{y-1}{x-1}&=3\\ y-1&=3(x-1)\\ y&=3x-2 \end{align*}
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