## Calculus with Applications (10th Edition)

$$\frac{{dy}}{{dx}} = \frac{{2y - 2\sqrt y }}{{4\sqrt y - x + 9y\sqrt y }}$$
\eqalign{ & \frac{{x + 2y}}{{x - 3y}} = {y^{1/2}} \cr & {\text{cross product}} \cr & x + 2y = {y^{1/2}}\left( {x - 3y} \right) \cr & x + 2y = x{y^{1/2}} - 3{y^{3/2}} \cr & {\text{take the derivative on both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {x + 2y} \right) = \frac{d}{{dx}}\left( {x{y^{1/2}} - 3{y^{3/2}}} \right) \cr & {\text{sum rule for derivatives}} \cr & \frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( {2y} \right) = \frac{d}{{dx}}\left( {x{y^{1/2}}} \right) - \frac{d}{{dx}}\left( {3{y^{3/2}}} \right) \cr & {\text{use the product rule for }}\frac{d}{{dx}}\left( {x{y^{1/2}}} \right) \cr & \frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( {2y} \right) = x\frac{d}{{dx}}\left( {{y^{1/2}}} \right) + {y^{1/2}}\frac{d}{{dx}}\left( x \right) - \frac{d}{{dx}}\left( {3{y^{3/2}}} \right) \cr & {\text{solve the derivatives using the chain rule}} \cr & 1 + 2\frac{{dy}}{{dx}} = x\left( {\frac{1}{{2\sqrt y }}} \right)\frac{{dy}}{{dx}} + {y^{1/2}}\left( 1 \right) - 3\left( {\frac{3}{2}} \right)y\frac{{dy}}{{dx}} \cr & 1 + 2\frac{{dy}}{{dx}} = \frac{x}{{2\sqrt y }}\frac{{dy}}{{dx}} + \sqrt y - \frac{{9y}}{2}\frac{{dy}}{{dx}} \cr & 2\sqrt y + 4\sqrt y \frac{{dy}}{{dx}} = x\frac{{dy}}{{dx}} + 2y - 9y\sqrt y \frac{{dy}}{{dx}} \cr & {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr & 4\sqrt y \frac{{dy}}{{dx}} - x\frac{{dy}}{{dx}} + 9y\sqrt y \frac{{dy}}{{dx}} = 2y - 2\sqrt y \cr & \left( {4\sqrt y - x + 9y\sqrt y } \right)\frac{{dy}}{{dx}} = 2y - 2\sqrt y \cr & \frac{{dy}}{{dx}} = \frac{{2y - 2\sqrt y }}{{4\sqrt y - x + 9y\sqrt y }} \cr}