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

Jan 26, 2016

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

#### Explanation:

The derivative of the whole function is

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

This will be more manageable if we split it up into each individual part first.

d/dx(y)=ul(color(blue)(dy/dx

The next is a little trickier. We should recognize that chain rule will be necessary, since $\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = u ' \cdot {e}^{u}$. Thus,

$\frac{d}{\mathrm{dx}} \left({e}^{x y}\right) = \frac{d}{\mathrm{dx}} \left(x y\right) \cdot {e}^{x y}$

To find $\frac{d}{\mathrm{dx}} \left(x y\right)$, use the product rule.

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

Plug this back in to the previous expression to see that

d/dx(e^(xy))=ul(color(blue)((y+xdy/dx)e^(xy)

For this last part, use the product rule once more.

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

To find the derivative of ${y}^{2}$, use the chain rule:

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

Which gives the derivative of the final part:

d/dx(-xy^2)=ul(color(blue)(-y^2-2xydy/dx

Plug these all back in for the derivative of the entire equation:

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

Now, solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$ through some algebraic manipulation.

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

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

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