# What is the slope of the tangent line of e^(xy)/(2x-y)= C , where C is an arbitrary constant, at (2,3)?

Jan 13, 2016

I would use the fact that $C$ is not arbitrary, but is determined by the fact that $\left(2 , 3\right)$ lies on the curve.

#### Explanation:

Since $\left(2 , 3\right)$ lies on the curve, we can find $C = {e}^{6}$.

The equation becomes:

${e}^{x y} = 2 {e}^{6} x - {e}^{6} y$     (for $y \ne 2 x$)

Differentiating implicitly yields:

${e}^{x y} \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 2 {e}^{6} - {e}^{6} \frac{\mathrm{dy}}{\mathrm{dx}}$

We were not asked for an explicit formula for $\frac{\mathrm{dy}}{\mathrm{dx}}$, so let's not bother with that.

At the point of interest $x = 2$ and $y = 3$, so we get:

${e}^{6} \left(3 + 2 \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 2 {e}^{6} - {e}^{6} \frac{\mathrm{dy}}{\mathrm{dx}}$

So we need only solve:

$3 + 2 \frac{\mathrm{dy}}{\mathrm{dx}} = 2 - \frac{\mathrm{dy}}{\mathrm{dx}}$ for $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{3}$ and we're done.