Question

What is the slope of the tangent line of `x/y-x^2y+e^y= C`, where C is an arbitrary constant, at `(-1,-1)`?

Answer 1

3e

Explanation:

The simplest approach is the Implicit Function Theorem

`dy/dx = - f_x / f_y`

here `f_x = 1/y - 2xy`

and `f_y = - x/y^2 - x^2 + e^y`

so `dy/dx = (y - 2xy^3)/ ( x + x^2 y^2 - y^2e^y)`

at -1, -1, that's `(-1 - (2)(-1)(-1))/ (-1 + 1 - e^{-1}) = 3e`