# How do you implicitly differentiate xy^2 + xy = 12?

Dec 24, 2015

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

#### Explanation:

Given $x {y}^{2} + x y = 12$

We can differentiate with respect to $\frac{d}{\mathrm{dx}}$ like this

$\frac{d}{\mathrm{dx}} \left(x {y}^{2}\right) + \frac{d}{\mathrm{dx}} \left(x y\right) = \frac{d}{\mathrm{dx}} \left(12\right)$

This can be done by product rule like this

$\frac{d}{\mathrm{dx}} \left(x\right) \left({y}^{2}\right) + x \cdot \frac{d}{\mathrm{dx}} \left({y}^{2}\right) + \frac{d}{\mathrm{dx}} \left(x\right) \cdot y + x \cdot \frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{\mathrm{dx}} \left(12\right)$

$\implies {y}^{2} + 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} + y + x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

Keep all $\frac{\mathrm{dy}}{\mathrm{dx}}$ on one side like this

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

Factor out $\frac{\mathrm{dy}}{\mathrm{dx}}$

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

Divide both side by $2 x y + x$ to get $\frac{\mathrm{dy}}{\mathrm{dx}}$ by itself

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

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