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

y'=e^(2y)/(x(2e^(2y)-x)

#### Explanation:

Start differentiating both sides of the equation with respect to $x$

$2 = y - \frac{{e}^{2 y}}{x}$

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

$0 = y ' - \frac{x \cdot \frac{d}{\mathrm{dx}} \left({e}^{2 y}\right) - {e}^{2 y} \cdot \frac{d}{\mathrm{dx}} \left(x\right)}{x} ^ 2$

$0 = y ' - \frac{x \cdot \left({e}^{2 y}\right) \cdot 2 y ' - {e}^{2 y} \cdot 1}{x} ^ 2$

$0 = y ' - \frac{2 {e}^{2 y} \cdot y '}{x} + {e}^{2 y} / {x}^{2}$

Transpose the terms with $y '$ to the left side of the equation

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

Factor out $y '$

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

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

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

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

y'=e^(2y)/(x(2e^(2y)-x)

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