# How do you differentiate 4e^y(x-y^2)=2(x-y)?

Dec 2, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 2 {e}^{y}}{1 + x - y - 4 y {e}^{y}}$

#### Explanation:

$4 x {e}^{y} - 4 {y}^{2} {e}^{y} = 2 x - 2 y$

$\frac{d}{\mathrm{dx}} \left[4 x {e}^{y} - 4 {y}^{2} {e}^{y} = 2 x - 2 y\right]$

${e}^{y} \frac{d}{\mathrm{dx}} \left[4 x\right] + 4 x \frac{d}{\mathrm{dx}} \left[{e}^{y}\right] - {e}^{y} \frac{d}{\mathrm{dx}} \left[4 {y}^{2}\right] - 4 {y}^{2} \frac{d}{\mathrm{dx}} \left[{e}^{y}\right] = \frac{d}{\mathrm{dx}} \left[2 x\right] - \frac{d}{\mathrm{dx}} \left[2 y\right]$

$4 {e}^{y} + 4 x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} - 8 y {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} - 4 {e}^{y} {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 2 - 2 \frac{\mathrm{dy}}{\mathrm{dx}}$

$4 x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} - 8 y {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} - 4 {e}^{y} {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 \frac{\mathrm{dy}}{\mathrm{dx}} = 2 - 4 {e}^{y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(4 x {e}^{y} - 8 y {e}^{y} - 4 {e}^{y} {y}^{2} + 2\right) = 2 - 4 {e}^{y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 - 4 {e}^{y}}{\textcolor{b l u e}{4 x {e}^{y}} - 8 y {e}^{y} \textcolor{b l u e}{- 4 {e}^{y} {y}^{2}} + 2}$

Notice how the terms in blue resemble the original equation. We can replace those terms with what they're equal to for a simpler derivative: color(blue)(2x-2y

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 - 4 {e}^{y}}{- 8 y {e}^{y} + \textcolor{b l u e}{2 x - 2 y} + 2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 2 {e}^{y}}{1 + x - y - 4 y {e}^{y}}$