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

Jul 21, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} y - 2 x y}{8 y + {x}^{2} - {x}^{3} - 1}$

#### Explanation:

Let's walk through this term by term.

First term, y is a function of x, using the fact that $\frac{d}{\mathrm{dx}} \left(y\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$ and the chain rule we obtain that

$\frac{d}{\mathrm{dx}} \left(4 {y}^{2}\right) = 8 y \frac{\mathrm{dy}}{\mathrm{dx}}$

Second term, we need to use the product rule:

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

Third:

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$

Fourth:

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

Combining these gives:

$8 y \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} y + {x}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x y - {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$

Gather like terms:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(8 y + {x}^{2} - {x}^{3} - 1\right) = 3 {x}^{2} y - 2 x y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} y - 2 x y}{8 y + {x}^{2} - {x}^{3} - 1}$