Question

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

Answer 1

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

Explanation:

Since `(2,3)` lies on the curve, we can find `C = e^6`.

The equation becomes:

`e^(xy)= 2e^6x-e^6y` `` `` (for `y != 2x`)

Differentiating implicitly yields:

`e^(xy)(y+x dy/dx)=2e^6-e^6 dy/dx`

We were not asked for an explicit formula for `dy/dx`, so let's not bother with that.

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

`e^6(3+2 dy/dx)= 2e^6-e^6 dy/dx`

So we need only solve:

`3+2 dy/dx = 2-dy/dx` for `dy/dx = -1/3` and we're done.