# How do you differentiate e^x/y=4x-y?

Oct 13, 2015

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

#### Explanation:

From ${e}^{x} / y = 4 x - y$ we have that

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

So differentiation both sides,

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

Using the product rule

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

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

Using the chain rule $\frac{d}{\mathrm{dx}} = \frac{d}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

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

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

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

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

However, as ${e}^{x} = 4 x y - {y}^{2}$, you might want to say it as

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