# What is the implicit derivative of 4= xy^2-y^3+xy ?

Oct 26, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y \left(1 + y\right)}{x + 2 x y - 3 {y}^{2}}$

#### Explanation:

Formally is quite easy. Calling $f \left(x , y\right) = x {y}^{2} - {y}^{3} + x y - 4 = 0$ we have:

$\mathrm{df} = {f}_{x} \mathrm{dx} + {f}_{y} \mathrm{dy} = 0$ so

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {f}_{x} / \left({f}_{y}\right) = - \frac{y \left(1 + y\right)}{x + 2 x y - 3 {y}^{2}}$

but suppose you need $\frac{\mathrm{dy}}{\mathrm{dx}}$ at ${x}_{0} = 1$. In this case you have to solve first

$f \left(1 , y\right) = {y}^{2} - {y}^{3} + y - 4 = 0$ giving

y_0=1/3 (1 - 4 (2/(97 - 9 sqrt[113]))^(1/3) - (1/2 (97 - 9 sqrt[113]))^( 1/3)) = -1.48558
as the real solution, and then

dy/dx)_(x_0,y_0) = 0.0839585