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

Nov 4, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2e-2}{4 - e}$

#### Explanation:

First find the derivative of the function and then plug in 1 for x and 1 for y
${\left(x + 2 y\right)}^{2} / \left(1 - {e}^{y}\right) = C$

Let's find the derivative of the function with respect to x and hold y constant.

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

Then find the derivative of the function with respect to y and hold x constant

Use the product rule ${f}_{y} = \frac{g f ' - f g '}{g} ^ 2$

${f}_{y} = \frac{\left(1 - {e}^{y}\right) \cdot 4 \left(x + 2 y\right) - {\left(x + 2 y\right)}^{2} \left(- {e}^{y}\right)}{1 - {e}^{y}} ^ 2$

${f}_{y} = \frac{\left(x + 2 y\right) \left[4 \left(1 - {e}^{y}\right) + {e}^{y} \left(x + 2 y\right)\right]}{1 - {e}^{y}} ^ 2$

Now to find the slope $\frac{\mathrm{dy}}{\mathrm{dx}}$ use the formula $\frac{\mathrm{dy}}{\mathrm{dx}} = - {f}_{x} / {f}_{y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \left({f}_{x} / {f}_{y}\right) = - \frac{\frac{2 \left(x + 2 y\right)}{1 - {e}^{y}}}{\frac{\left(x + 2 y\right) \left[4 \left(1 - {e}^{y}\right) + {e}^{y} \left(x + 2 y\right)\right]}{1 - {e}^{y}} ^ 2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \left({f}_{x} / {f}_{y}\right) = - \left(\frac{2 \left(\cancel{x + 2 y}\right)}{\cancel{1 - {e}^{y}}}\right) \cdot \frac{{\left(1 - {e}^{y}\right)}^{\cancel{2}}}{\cancel{\left(x + 2 y\right)} \left[4 \left(1 - {e}^{y}\right) + {e}^{y} \left(x + 2 y\right)\right]}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \left({f}_{x} / {f}_{y}\right) = \frac{- 2 + 2 {e}^{y}}{4 - 4 {e}^{y} + x {e}^{y} + 2 y {e}^{y}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \left({f}_{x} / {f}_{y}\right) = \frac{- 2 + 2 {e}^{1}}{4 - 4 {e}^{1} + \left(1\right) {e}^{1} + 2 \left(1\right) {e}^{1}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2e-2}{4 - e}$