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

$y ' = \frac{{x}^{2} \cdot {e}^{2 x - 4 y} + y}{2 {x}^{2} \cdot {e}^{2 x - 4 y} + x - {x}^{2} y}$

#### Explanation:

Start differentiating with respect to x both sides of the equation
$- {y}^{2} = {e}^{2 x - 4 y} - 2 \cdot \frac{y}{x}$

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

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

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

Simplify by dividing by 2

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

$- y \cdot y ' = {e}^{2 x - 4 y} - 2 \cdot {e}^{2 x - 4 y} y ' - \frac{x y '}{x} ^ 2 + \frac{y}{x} ^ 2$

$- y \cdot y ' = {e}^{2 x - 4 y} - 2 \cdot {e}^{2 x - 4 y} y ' - \frac{y '}{x} + \frac{y}{x} ^ 2$

$2 \cdot {e}^{2 x - 4 y} y ' + \frac{y '}{x} - y \cdot y ' = {e}^{2 x - 4 y} + \frac{y}{x} ^ 2$

Factor the y'

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

Divide both sides by $\left(2 \cdot {e}^{2 x - 4 y} + \frac{1}{x} - y\right)$

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

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

$y ' = \frac{{e}^{2 x - 4 y} + \frac{y}{x} ^ 2}{2 \cdot {e}^{2 x - 4 y} + \frac{1}{x} - y}$

Simplify by multiplying the right side by ${x}^{2} / {x}^{2}$

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

$y ' = \frac{{x}^{2} \cdot {e}^{2 x - 4 y} + y}{2 {x}^{2} \cdot {e}^{2 x - 4 y} + x - {x}^{2} y}$

God bless....I hope the explanation is useful.