# What is the slope of the tangent line of e^(xy)-e^x/y = C , where C is an arbitrary constant, at (0,1)?

Sep 19, 2017

The slope is $m = 0$

#### Explanation:

If the curve goes through the point $\left(0 , 1\right)$ then we have:

${e}^{0 \times 1} - {e}^{0} / 1 = C$

$1 - 1 = C$

$C = 0$

So the only curve of the family going through that point would be:

${e}^{x y} = {e}^{x} / y$

Taking the logarithm of both sides we have:

$x y = x - \ln y$

Now if $y = 1$ this equation becomes $x = x$, which is satisfied by any value of $x$

If $y \ne 1$ we have:

$x = \ln \frac{y}{1 - y} \ne 0$

So the curve passing through the point $\left(0 , 1\right)$ is the straight line $y = 1$, and the tangent is the horizontal line itself with slope $m = 0$.