# 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)?

Jun 21, 2016

3e

#### Explanation:

The simplest approach is the Implicit Function Theorem

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {f}_{x} / {f}_{y}$

here ${f}_{x} = \frac{1}{y} - 2 x y$

and ${f}_{y} = - \frac{x}{y} ^ 2 - {x}^{2} + {e}^{y}$

so $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y - 2 x {y}^{3}}{x + {x}^{2} {y}^{2} - {y}^{2} {e}^{y}}$

at -1, -1, that's $\frac{- 1 - \left(2\right) \left(- 1\right) \left(- 1\right)}{- 1 + 1 - {e}^{- 1}} = 3 e$