## Calculus with Applications (10th Edition)

$$\frac{{dy}}{{dx}} = \frac{{2x - 9{x^2}{y^4}}}{{12{x^3}{y^3} + 8y}}$$
\eqalign{ & {x^2} - 4{y^2} = 3{x^3}{y^4} \cr & {\text{take the derivative on both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {{x^2} - 4{y^2}} \right) = \frac{d}{{dx}}\left( {3{x^3}{y^4}} \right) \cr & {\text{use sum rule and product rule for differentiation as follows}} \cr & \frac{d}{{dx}}\left( {{x^2}} \right) - \frac{d}{{dx}}\left( {4{y^2}} \right) = 3{x^3}\frac{d}{{dx}}\left( {{y^4}} \right) + {y^4}\frac{d}{{dx}}\left( {3{x^3}} \right) \cr & {\text{solve the derivatives using the power rule and the chain rule}} \cr & 2x - 8y\frac{{dy}}{{dx}} = 3{x^3}\left( {4{y^3}\frac{{dy}}{{dx}}} \right) + {y^4}\left( {9{x^2}} \right) \cr & 2x - 8y\frac{{dy}}{{dx}} = 12{x^3}{y^3}\frac{{dy}}{{dx}} + 9{x^2}{y^4} \cr & {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr & 2x - 9{x^2}{y^4} = 12{x^3}{y^3}\frac{{dy}}{{dx}} + 8y\frac{{dy}}{{dx}} \cr & 2x - 9{x^2}{y^4} = \left( {12{x^3}{y^3} + 8y} \right)\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = \frac{{2x - 9{x^2}{y^4}}}{{12{x^3}{y^3} + 8y}} \cr}