Chapter 6 - Applications of the Derivative - 6.4 Implicit Differentiation - 6.4 Exercises - Page 334: 13

$$\frac{{dy}}{{dx}} = \frac{{5 - 2xy{e^{{x^2}y}}}}{{{x^2}{e^{{x^2}y}} - 4}}$$

Work Step by Step

\eqalign{ & {e^{{x^2}y}} = 5x + 4y + 2 \cr & {\text{take the derivative on both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {{e^{{x^2}y}}} \right) = \frac{d}{{dx}}\left( {5x + 4y + 2} \right) \cr & {\text{use sum rule as follows}} \cr & \frac{d}{{dx}}\left( {{e^{{x^2}y}}} \right) = \frac{d}{{dx}}\left( {5x} \right) + \frac{d}{{dx}}\left( {4y} \right) + \frac{d}{{dx}}\left( 2 \right) \cr & {\text{solve the derivatives using the chain rule and the basic integration rules}} \cr & {e^{{x^2}y}}\frac{d}{{dx}}\left( {{x^2}y} \right) = 5 + 4\frac{{dy}}{{dx}} + 0 \cr & {\text{use the product rule}} \cr & {e^{{x^2}y}}\left[ {{x^2}\frac{d}{{dx}}\left( y \right) + y\frac{d}{{dx}}\left( {{x^2}} \right)} \right] = 5 + 4\frac{{dy}}{{dx}} \cr & {e^{{x^2}y}}\left[ {{x^2}\frac{{dy}}{{dx}} + 2xy} \right] = 5 + 4\frac{{dy}}{{dx}} \cr & {x^2}{e^{{x^2}y}}\frac{{dy}}{{dx}} + 2xy{e^{{x^2}y}} = 5 + 4\frac{{dy}}{{dx}} \cr & {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr & {x^2}{e^{{x^2}y}}\frac{{dy}}{{dx}} - 4\frac{{dy}}{{dx}} = 5 - 2xy{e^{{x^2}y}} \cr & \left( {{x^2}{e^{{x^2}y}} - 4} \right)\frac{{dy}}{{dx}} = 5 - 2xy{e^{{x^2}y}} \cr & \frac{{dy}}{{dx}} = \frac{{5 - 2xy{e^{{x^2}y}}}}{{{x^2}{e^{{x^2}y}} - 4}} \cr}

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