# How do you differentiate e^(xy)+x^2-y^2=0?

Mar 9, 2015

You have an Implicit Function in which $y$ is function of $x$ so you have to derive it as well.
You get:
${e}^{x y} \cdot \left(1 \cdot y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) + 2 x - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
where you used the Product Rule for the exponent of $e$;

$y {e}^{x y} + x {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left[x {e}^{x y} - 2 y\right] = - 2 x - y {e}^{x y}$

finally:

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

Hope it helps