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

Jul 2, 2017

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

#### Explanation:

Implicit differentiation is a special case of the chain rule for derivatives. Generally differentiation problems involve functions i.e. $y = f \left(x\right)$ - written explicitly as functions of $x$. However, some functions y are written implicitly as functions of $x$. So what we do is to treat $y$ as $y = y \left(x\right)$ and use chain rule. This means differentiating $y$ w.r.t. $y$, but as we have to derive w.r.t. $x$, as per chain rule, we multiply it by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

Here we have $y = \left(x - {y}^{2}\right) {e}^{x y} - x y$

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

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

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

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