# What is the slope of the tangent line of sqrt(ye^(x-y))= C , where C is an arbitrary constant, at (-2,1)?

$s l o p e = - \frac{1}{0} = - \infty$

#### Explanation:

the given $\sqrt{y \cdot {e}^{x - y}} = C$ at $\left(- 2 , 1\right)$

squaring both sides of the equation it becomes

$y \cdot {e}^{x - y} = {c}^{2}$

differentiation both sides of the equation

$y ' {e}^{x - y} + y \cdot \left({e}^{x - y}\right) \left(1 - y '\right) = 0$

$y ' {e}^{x - y} + y {e}^{x - y} - y {e}^{x - y} \cdot y ' = 0$

$y ' \left({e}^{x - y} - y {e}^{x - y}\right) = - y {e}^{x - y}$

$y ' = \frac{- y \left({e}^{x - y}\right)}{\left({e}^{x - y}\right) \left(1 - y\right)}$

$y ' = - \frac{y}{1 - y}$

Evaluating $y '$ at $y = 1$

$y ' = - \frac{1}{1 - 1} = - \frac{1}{0} = - \infty$