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: 33


$$ y = -2x+2$$

Work Step by Step

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