# How do you use implicit differentiation to find dy/dx given 2e^(3x)+e^(-2y)+7=0?

Nov 2, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 {e}^{3 x}}{2 {e}^{- 2 y}} = \frac{3 {e}^{3 x}}{e} ^ \left(- 2 y\right)$

#### Explanation:

$2 {e}^{3 x} + {e}^{- 2 y} + 7 = 0$

$6 {e}^{3 x} - 2 {e}^{- 2 y} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$6 {e}^{3 x} = 2 {e}^{- 2 y} \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 {e}^{3 x}}{2 {e}^{- 2 y}} = \frac{3 {e}^{3 x}}{e} ^ \left(- 2 y\right)$