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: 32


$$y =-x+2$$

Work Step by Step

Given $$x^{2 / 3}+y^{2 / 3}=2,\ \ \ \ (1,1) $$ Differentiate both sides with respect to $x$ \begin{align*} \frac{d}{dx}(x^{2 / 3}+y^{2 / 3})&=\frac{d}{dx}(2)\\ \frac{2}{3} x^{-\frac{1}{5}}+\frac{2}{3} y^{-\frac{1}{5}} y^{\prime}&=0\\ \frac{2}{3} y^{-\frac{1}{3}} y^{\prime}&=-\frac{2}{3} x^{-\frac{1}{3}}\\ y'&= \frac{-x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}} \end{align*} Then at $(2,1)$ \begin{align*} m&= \frac{-1^{-\frac{1}{3}}}{1^{-\frac{1}{3}}}\\ &=- 1 \end{align*} Then the tangent line is given by \begin{align*} \frac{y-y_1}{x-x_1}&=m\\ \frac{y-1}{x-1}&=-1\\ y-1 &= -x+1\\ y&=-x+2 \end{align*}
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