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

Nov 29, 2016

(dy)/(dx)=(4-y)/[4-2y+x+2e^(2y

#### Explanation:

Firstly multiply everything by $\text{ "(y-x)" }$ to remove the award denominator.

assuming $y \ne x$

$4 \left(y - x\right) = y \left(y - x\right) - {e}^{2 y}$

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

now differentiate.

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

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

now rearrange.

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

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

(dy)/(dx)=(4-y)/[4-2y+x+2e^(2y