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

Dec 24, 2015

$\frac{{e}^{4} - 1}{1 + 4 {e}^{4}}$

#### Explanation:

Apply the product rule:

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

To find the derivative of ${e}^{2 y}$, the chain rule will be necessary.

$\frac{d}{\mathrm{dx}} \left({e}^{2 y}\right) = {e}^{2 y} \cdot \frac{d}{\mathrm{dx}} \left(2 y\right)$
$\implies {e}^{2 y} \cdot 2 \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

Plug this back into the differentiated equation and simplify the other derivatives:

$\frac{\mathrm{dy}}{\mathrm{dx}} {e}^{x} + y {e}^{x} - {e}^{2 y} - \frac{\mathrm{dy}}{\mathrm{dx}} 2 x {e}^{2 y} = 0$

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

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

Continued simplification yields:

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

To find the slope of the tangent line, plug in the $x$ and $y$ values $\left(- 2 , 1\right)$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{2 \times 1} - \left(1\right) {e}^{-} 2}{{e}^{-} 2 - 2 \left(- 2\right) {e}^{2 \times 1}}$

Simplify:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{4} - 1}{1 + 4 {e}^{4}} \approx 0.2443$

The slope could be written in either form—I prefer the more exact version, but knowing the decimal equivalence can be useful for checking your answer or for a continued application.