# How do you implicitly differentiate 5y^2=e^(x-4y)-3x ?

Oct 23, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x - 4 y} - 3}{10 + 4 {e}^{x - 4 y}}$

#### Explanation:

differentiate each part
$5 {y}^{2} = {e}^{x - 4 y} - 3 x$

$5 {y}^{2} \implies 10 y \frac{\mathrm{dy}}{\mathrm{dx}}$

${e}^{x - 4 y} \implies {e}^{x - 4 y} - 4 {e}^{x - 4 y} \frac{\mathrm{dy}}{\mathrm{dx}}$
(this part is the most difficult so I'll do it in a little more depth)

$y = {e}^{x - 4 y}$ using the rule $y = {e}^{f} \left(x\right) \implies f ' \left(x\right) {e}^{f} \left(x\right)$

giving us

$y ' = \left(1 - 4 \frac{\mathrm{dy}}{\mathrm{dx}}\right) {e}^{x - 4 y}$

$y ' {e}^{x - 4 y} - 4 {e}^{x - 4 y} \frac{\mathrm{dy}}{\mathrm{dx}}$

$- 3 x \implies - 3$

leaving us with

$10 y \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x - 4 y} - 4 {e}^{x - 4 y} \frac{\mathrm{dy}}{\mathrm{dx}} - 3$

$10 y \frac{\mathrm{dy}}{\mathrm{dx}} + 4 {e}^{x - 4 y} \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x - 4 y} - 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(10 y + 4 {e}^{x - 4 y}\right) = {e}^{x - 4 y} - 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x - 4 y} - 3}{10 y + 4 {e}^{x - 4 y}}$