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: 35


$$y = -\frac{12}{5} x+\frac{32}{5}$$

Work Step by Step

Given $$2 x^{1 / 2}+4 y^{-1 / 2}=x y,\ \ \ (1,4)$$ Differentiate both sides with respect to $x$ \begin{align*} \frac{d}{dx}(2 x^{1 / 2}+4 y^{-1 / 2} )&=\frac{d}{dx}(x y )\\ x^{-\frac{1}{2}}-2 y^{-\frac{3}{2}} y^{\prime}&=x y^{\prime}+y\\ \left(-2 y^{-\frac{3}{2}}-x\right) y^{\prime}&=y-x^{-\frac{1}{2}}\\ y^{\prime}&=\frac{y-x^{-\frac{1}{2}}}{-2 y^{-\frac{1}{2}}-x} \end{align*} Then at $(1,4)$ \begin{align*} m&= \frac{4-(1)^{-\frac{1}{2}}}{-2(4)^{-\frac{3}{2}}-1}\\ &= \frac{-12}{5} \end{align*} Then the tangent line is given by \begin{align*} \frac{y-y_1}{x-x_1}&=m\\ \frac{y-4}{x-1}&= \frac{-12}{5}\\ y-4 &=-\frac{12}{5} \cdot(x-1) \\ y &=-\frac{12}{5} x+\frac{12}{5}+4\\ y&= -\frac{12}{5} x+\frac{32}{5} \end{align*}
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