Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 6 - Applications of the Derivative - 6.4 Implicit Differentiation - 6.4 Exercises - Page 334: 21

Answer

$$y = \frac{1}{{64}}x + \frac{7}{4}$$

Work Step by Step

$$\eqalign{ & 2{y^2} - \sqrt x = 4;{\text{ }}\left( {16,2} \right) \cr & 2{y^2} - {x^{1/2}} = 4 \cr & {\text{take the derivative on both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {2{y^2} - {x^{1/2}}} \right) = \frac{d}{{dx}}\left( 4 \right) \cr & {\text{use sum rule as follows}} \cr & \frac{d}{{dx}}\left( {2{y^2}} \right) - \frac{d}{{dx}}\left( {{x^{1/2}}} \right) = \frac{d}{{dx}}\left( 4 \right) \cr & {\text{solve the derivatives using the power rule and the chain rule}} \cr & 4y\frac{{dy}}{{dx}} - \frac{1}{2}{x^{ - 1/2}} = 0 \cr & {\text{solving the equation for }}\frac{{dy}}{{dx}} \cr & 4y\frac{{dy}}{{dx}} = \frac{1}{{2{x^{1/2}}}} \cr & \frac{{dy}}{{dx}} = \frac{1}{{8{x^{1/2}}y}} \cr & \cr & {\text{find the slope at the given point }} \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {16,2} \right)}} = \frac{1}{{8{{\left( {16} \right)}^{1/2}}\left( 2 \right)}} = \frac{1}{{64}} \cr & {\text{find the equation of the tangent line at the point }}\left( {16,2} \right) \cr & {\text{by using the point - slope form }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 2 = \frac{1}{{64}}\left( {x - 16} \right) \cr & y - 2 = \frac{1}{{64}}x - \frac{1}{4} \cr & y = \frac{1}{{64}}x - \frac{1}{4} + 2 \cr & y = \frac{1}{{64}}x + \frac{7}{4} \cr} $$
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