# How do you differentiate x * y + 2x + 3x^2 = 4?

##### 1 Answer
Aug 3, 2015

$\frac{d}{\mathrm{dx}} \left[x y + 2 x + 3 {x}^{2} = 4\right]$ results in $\frac{\mathrm{dy}}{\mathrm{dx}} = 3 x - 4 - \frac{6}{x}$

#### Explanation:

Using Implicit Differentiation we have

$\frac{d}{\mathrm{dx}} \left[x y + 2 x + 3 {x}^{2} = 4\right]$
$\frac{d}{\mathrm{dx}} \left[x y\right] + \frac{d}{\mathrm{dx}} \left[2 x\right] + \frac{d}{\mathrm{dx}} \left[3 {x}^{2}\right] = 0$
$x \cdot \frac{d}{\mathrm{dx}} \left[y\right] + y \cdot \frac{d}{\mathrm{dx}} \left[x\right] + 2 \cdot \frac{d}{\mathrm{dx}} \left[x\right] + 3 \cdot \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] = 0$
$x \cdot y ' + y \cdot 1 + 2 \cdot 1 + 3 \cdot \left(2 x\right) = 0$
$x \cdot y ' + y = - 6 x - 2$
$y ' = \frac{- 6 x - 2}{x} - y$

Solving for y in the original problem, we have

$x \cdot y + 2 x + 3 {x}^{2} = 4$
$x \cdot y = 4 - 2 x - 3 {x}^{2}$
$y = \frac{4 - 2 x - 3 {x}^{2}}{x}$

Now, plug this value for $y$ into the equation for $y '$ above

$y ' = \frac{- 6 x - 2}{x} - y$
$y ' = \frac{- 6 x - 2}{x} - \frac{4 - 2 x - 3 {x}^{2}}{x}$
$y ' = \frac{- 6 x - 2 - 4 + 2 x + 3 {x}^{2}}{x}$
$y ' = \frac{- 4 x - 6 + 3 {x}^{2}}{x}$
$y ' = \frac{3 {x}^{2} - 4 x - 6}{x}$
$y ' = 3 x - 4 - \frac{6}{x}$