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

slope $= 2$

#### Explanation:

Start with the given $\left(\frac{y}{x}\right) {e}^{\frac{x}{y}} = C$ at $\left(1 , 2\right)$

obtain the derivative of both sides of the equation

$\left(\frac{y}{x}\right) {e}^{\frac{x}{y}} \cdot \left(\frac{y \cdot 1 - x \cdot y '}{y} ^ 2\right) + {e}^{\frac{x}{y}} \cdot \left(\frac{x y ' - y \cdot 1}{x} ^ 2\right) = 0$

After dividing both sides by ${e}^{\frac{x}{y}}$

$\left(\frac{y}{x}\right) \cdot \left(\frac{y \cdot 1 - x \cdot y '}{y} ^ 2\right) + \left(\frac{x y ' - y \cdot 1}{x} ^ 2\right) = 0$

$- y \frac{'}{y} + y \frac{'}{x} = \frac{y}{x} ^ 2 - \frac{1}{x}$

$y ' \left(\frac{1}{x} - \frac{1}{y}\right) = \frac{y}{x} ^ 2 - \frac{1}{x}$

$y ' = \frac{\frac{y}{x} ^ 2 - \frac{1}{x}}{\frac{1}{x} - \frac{1}{y}}$

substitute now the values $x = 1$ and $y = 2$

$y ' = \frac{\frac{2}{1} ^ 2 - \frac{1}{1}}{\frac{1}{1} - \frac{1}{2}} = \frac{2 - 1}{1 - \frac{1}{2}} = 2$