# How do you implicitly differentiate 7=3x-y+y^2x?

Dec 17, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{2} + 3}{1 - 2 x y}$

#### Explanation:

The challenge in this problem lies in using the product rule for the ${y}^{2} x$ term.

$\frac{d}{\mathrm{dx}} \left[7 = 3 x - y + {y}^{2} x\right]$

$0 = 3 - \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{d}{\mathrm{dx}} \left[{y}^{2}\right] \cdot x + \frac{d}{\mathrm{dx}} \left[x\right] \cdot {y}^{2}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} - 2 x y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2} + 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(1 - 2 x y\right) = {y}^{2} + 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{2} + 3}{1 - 2 x y}$