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

Mar 19, 2018

One form of the answer is $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{8 {e}^{2} + 13}{5 - 2 {e}^{2}}$

#### Explanation:

Implicitly differentiate

$\left(2 x - \frac{\mathrm{dy}}{\mathrm{dx}}\right) {e}^{y} + \left({x}^{2} - y\right) {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\frac{\mathrm{dy}}{\mathrm{dx}} x - y}{x} ^ 2 + \frac{3 {x}^{2} y - {x}^{3} \frac{\mathrm{dy}}{\mathrm{dx}}}{y} ^ 2 = 0$

Collect derivative terms

$- \frac{\mathrm{dy}}{\mathrm{dx}} \left[{e}^{y} \left(1 - {x}^{2} + y\right) + \frac{1}{x} + {x}^{3} / {y}^{2}\right] + 2 x {e}^{y} + \frac{y}{x} ^ 2 + 3 {x}^{2} / y = 0$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x {e}^{y} + \frac{y}{x} ^ 2 + 3 {x}^{2} / y}{{e}^{y} \left(1 - {x}^{2} + y\right) + \frac{1}{x} + {x}^{3} / {y}^{2}}$

Substitute in the values $x = 2$ and $y = 2$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \cdot 2 {e}^{2} + \frac{2}{2} ^ 2 + 3 \cdot {2}^{2} / 2}{{e}^{2} \left(1 - {2}^{2} + 2\right) + \frac{1}{2} + {2}^{3} / {2}^{2}} = \frac{4 {e}^{2} + \frac{1}{2} + 6}{- {e}^{2} + \frac{1}{2} + 2}$

This will look a little simpler if we multiply the numerator and denominator by 2.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{8 {e}^{2} + 13}{5 - 2 {e}^{2}}$