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: 37


$$y = 2x+\pi$$

Work Step by Step

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