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: 31


$$y = \frac{-1}{2}x+2$$

Work Step by Step

Given $$x y+x^{2} y^{2}=6,\ \ \ \ (2,1) $$ Differentiate both sides with respect to $x$ \begin{align*} \frac{d}{dx}(x y+x^{2} y^{2})&=\frac{d}{dx}(6)\\ x y^{\prime}+y+2 x y^{2}+2 x^{2} y y^{\prime}&=0\\ [x+2x^2y]y'&= -y-2 x y^{2} \end{align*} Then at $(2,1)$ \begin{align*} m&= \frac{ -1-2 (2)}{2+2(4)}\\ &= \frac{-1}{2} \end{align*} Then the tangent line is given by \begin{align*} \frac{y-y_1}{x-x_1}&=m\\ \frac{y-1}{x-2}&=\frac{-1}{2}\\ 2(y-1)&= -x+2\\ y-1&= \frac{-1}{2}x+1\\ y&= \frac{-1}{2}x+2 \end{align*}
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