# How do you differentiate x^4(x+y)=y^2(3x-y)?

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {y}^{2} - 5 {x}^{4} - 4 {x}^{3} y}{{x}^{4} - 6 x y + 3 {y}^{2}}$
This can be by implicit differentiation. Rewrite it as ${x}^{5} + {x}^{4} y = 3 x {y}^{2} - {y}^{3}$. Now differentiate term wise, $5 {x}^{4} + {x}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} + 4 {x}^{3} y = 3 {y}^{2} + 6 x y \frac{\mathrm{dy}}{\mathrm{dx}} - 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$
$\left({x}^{4} - 6 x y + 3 {y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {y}^{2} - 5 {x}^{4} - 4 {x}^{3} y$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {y}^{2} - 5 {x}^{4} - 4 {x}^{3} y}{{x}^{4} - 6 x y + 3 {y}^{2}}$