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

Feb 15, 2016

dy/dx=(3x^2-e^(x-xy)+ye^(x-xy)-2y)/(8y+2x+xe(x-xy)

#### Explanation:

Since it is not possible to write $y$ as an explicit function of $x$, we use the method of implicit differentiation.

This entails differentiating both sides of the equation with respect to $x$, bearing in mind that $y$ is a function of $x$, and then solving for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

Doing so yields :

$\frac{d}{\mathrm{dx}} \left({x}^{3} - {e}^{x - x y} - 4 {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(2 x y\right)$

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

thereforedy/dx=(3x^2-e^(x-xy)+ye^(x-xy)-2y)/(8y+2x+xe(x-xy)