## Calculus with Applications (10th Edition)

$$\frac{{dy}}{{dx}} = - \frac{{6{x^{1/2}} + 4{x^3}{y^3}}}{{3{x^4}{y^2} - 9{y^{1/2}}}}$$
\eqalign{ & {x^4}{y^3} + 4{x^{3/2}} = 6{y^{3/2}} + 5 \cr & {\text{take the derivative on both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {{x^4}{y^3} + 4{x^{3/2}}} \right) = \frac{d}{{dx}}\left( {6{y^{3/2}} + 5} \right) \cr & {\text{use sum rule and product rule for differentiation as follows}} \cr & \frac{d}{{dx}}\left( {{x^4}{y^3}} \right) + \frac{d}{{dx}}\left( {4{x^{3/2}}} \right) = \frac{d}{{dx}}\left( {6{y^{3/2}}} \right) + \frac{d}{{dx}}\left( 5 \right) \cr & {y^3}\frac{d}{{dx}}\left( {{x^4}} \right) + {x^4}\frac{d}{{dx}}\left( {{y^3}} \right) + \frac{d}{{dx}}\left( {4{x^{3/2}}} \right) = \frac{d}{{dx}}\left( {6{y^{3/2}}} \right) + \frac{d}{{dx}}\left( 5 \right) \cr & {\text{solve the derivatives using the power rule and the chain rule}} \cr & {y^3}\left( {4{x^3}} \right) + {x^4}\left( {3{y^2}} \right)\frac{{dy}}{{dx}} + 4\left( {\frac{3}{2}} \right){x^{1/2}} = 6\left( {\frac{3}{2}} \right){y^{1/2}}\frac{{dy}}{{dx}} + 0 \cr & 4{x^3}{y^3} + 3{x^4}{y^2}\frac{{dy}}{{dx}} + 6{x^{1/2}} = 9{y^{1/2}}\frac{{dy}}{{dx}} \cr & {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr & 3{x^4}{y^2}\frac{{dy}}{{dx}} - 9{y^{1/2}}\frac{{dy}}{{dx}} = - 6{x^{1/2}} - 4{x^3}{y^3} \cr & \left( {3{x^4}{y^2} - 9{y^{1/2}}} \right)\frac{{dy}}{{dx}} = - 6{x^{1/2}} - 4{x^3}{y^3} \cr & \frac{{dy}}{{dx}} = \frac{{ - 6{x^{1/2}} - 4{x^3}{y^3}}}{{3{x^4}{y^2} - 9{y^{1/2}}}} \cr & \frac{{dy}}{{dx}} = - \frac{{6{x^{1/2}} + 4{x^3}{y^3}}}{{3{x^4}{y^2} - 9{y^{1/2}}}} \cr}