## Calculus with Applications (10th Edition)

$$\frac{{dy}}{{dx}} = \frac{{3{x^{3/2}}{y^{5/2}} - 2y}}{{2x\ln x - 5{x^{5/2}}{y^{3/2}}}}$$
\eqalign{ & y\ln x + 2 = {x^{3/2}}{y^{5/2}} \cr & {\text{take the derivative on both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {y\ln x + 2} \right) = \frac{d}{{dx}}\left( {{x^{3/2}}{y^{5/2}}} \right) \cr & {\text{use sum rule as follows}} \cr & \frac{d}{{dx}}\left( {y\ln x} \right) + \frac{d}{{dx}}\left( 2 \right) = \frac{d}{{dx}}\left( {{x^{3/2}}{y^{5/2}}} \right) \cr & {\text{use the product rule}} \cr & y\frac{d}{{dx}}\left( {\ln x} \right) + \ln x\frac{d}{{dx}}\left( y \right) + \frac{d}{{dx}}\left( 2 \right) = {x^{3/2}}\frac{d}{{dx}}\left( {{y^{5/2}}} \right) + {y^{5/2}}\frac{d}{{dx}}\left( {{x^{3/2}}} \right) \cr & {\text{solve the derivatives}} \cr & y\left( {\frac{1}{x}} \right) + \ln x\frac{{dy}}{{dx}} + 0 = {x^{3/2}}\left( {\frac{5}{2}{y^{3/2}}} \right)\frac{{dy}}{{dx}} + {y^{5/2}}\left( {\frac{3}{2}{x^{1/2}}} \right) \cr & \frac{y}{x} + \ln x\frac{{dy}}{{dx}} = \frac{5}{2}{x^{3/2}}{y^{3/2}}\frac{{dy}}{{dx}} + \frac{3}{2}{x^{1/2}}{y^{5/2}} \cr & \ln x\frac{{dy}}{{dx}} - \frac{5}{2}{x^{3/2}}{y^{3/2}}\frac{{dy}}{{dx}} = \frac{3}{2}{x^{1/2}}{y^{5/2}} - \frac{y}{x} \cr & {\text{multiply both sides by }}2x \cr & 2x\ln x\frac{{dy}}{{dx}} - 5{x^{5/2}}{y^{3/2}}\frac{{dy}}{{dx}} = 3{x^{3/2}}{y^{5/2}} - 2y \cr & {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr & \left( {2x\ln x - 5{x^{5/2}}{y^{3/2}}} \right)\frac{{dy}}{{dx}} = 3{x^{3/2}}{y^{5/2}} - 2y \cr & \frac{{dy}}{{dx}} = \frac{{3{x^{3/2}}{y^{5/2}} - 2y}}{{2x\ln x - 5{x^{5/2}}{y^{3/2}}}} \cr}