# How do you implicitly differentiate 9=ye^(2y)/x?

Jan 6, 2016

$\frac{y}{x + 2 x y} = \frac{\mathrm{dy}}{\mathrm{dx}}$

#### Explanation:

First pick one to be dependent on the other, for convenience and tradition's sake I'll say that $x$ depends on $y$.

$9 = y \frac{{e}^{2 y}}{x}$

Differentiate both sides

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

Use the product rule

$0 = \frac{{e}^{2 y}}{x} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + y \frac{d}{\mathrm{dx}} \frac{{e}^{2 y}}{x}$

And again (or use the quotient rule)

$0 = \frac{{e}^{2 y}}{x} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + y \left(\frac{1}{x} \cdot \frac{d}{\mathrm{dx}} \left({e}^{2 y}\right) + {e}^{2 y} \frac{d}{\mathrm{dx}} \cdot \frac{1}{x}\right)$
$0 = \frac{{e}^{2 y}}{x} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + y \left(\frac{1}{x} \cdot \frac{d}{\mathrm{dx}} \left({e}^{2 y}\right) - {e}^{2 y} / {x}^{2}\right)$

Say that $2 y = u$

$0 = \frac{{e}^{2 y}}{x} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + y \left(\frac{1}{x} \cdot \frac{d}{\mathrm{du}} {e}^{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}} - {e}^{2 y} / {x}^{2}\right)$
$0 = \frac{{e}^{2 y}}{x} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + y \left({e}^{2 y} / x \cdot \frac{\mathrm{du}}{\mathrm{dx}} - {e}^{2 y} / {x}^{2}\right)$

Since $2 y = u$, $\frac{\mathrm{du}}{\mathrm{dx}} = 2 \frac{\mathrm{dy}}{\mathrm{dx}}$
$0 = \frac{{e}^{2 y}}{x} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y {e}^{2 y} / x \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - y {e}^{2 y} / {x}^{2}$

But since $9 = y {e}^{2 y} / x$

$0 = \frac{9}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + 18 \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{9}{x}$

Put $\frac{\mathrm{dy}}{\mathrm{dx}}$ in evidence

$\frac{9}{x} = \frac{\mathrm{dy}}{\mathrm{dx}} \left(\frac{9}{y} + 18\right)$

Pass it over dividing

$\frac{9}{x \left(\frac{9}{y} + 18\right)} = \frac{\mathrm{dy}}{\mathrm{dx}}$

Use algebra to make it look prettier if you wish

$\frac{9}{9 \frac{x}{y} + 18 x} = \frac{\mathrm{dy}}{\mathrm{dx}}$
$\frac{1}{\frac{x}{y} + 2 x} = \frac{\mathrm{dy}}{\mathrm{dx}}$
$\frac{y}{x + 2 x y} = \frac{\mathrm{dy}}{\mathrm{dx}}$

If you need $\frac{\mathrm{dx}}{\mathrm{dy}}$ just take the inverse of $\frac{\mathrm{dy}}{\mathrm{dx}}$