# How do you find dy/dx implicitly of ye^(3x) - 4^x = 2y^2?

May 6, 2015

$y {e}^{3 x} - {4}^{x} = 2 {y}^{2}$

To differentiate we'll need the product rule for the first term.

$\frac{d}{\mathrm{dx}} \left(y {e}^{3 x} - {4}^{x}\right) = \frac{d}{\mathrm{dx}} \left(2 {y}^{2}\right)$

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

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

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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