# How do you implicitly differentiate  xy+2x+3x^2=-4?

May 13, 2018

So, recall that for implicit differentiation, each term has to be differentiated with respect to a single variable, and that to differentiate some $f \left(y\right)$ with respect to $x$, we utilise the chain rule:

$\frac{d}{\mathrm{dx}} \left(f \left(y\right)\right) = f ' \left(y\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

Thus, we state the equality:
$\frac{d}{\mathrm{dx}} \left(x y\right) + \frac{d}{\mathrm{dx}} \left(2 x\right) + \frac{d}{\mathrm{dx}} \left(3 {x}^{2}\right) = \frac{d}{\mathrm{dx}} \left(- 4\right)$
$\Rightarrow x \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + y + 2 + 6 x = 0$ (using the product rule to differentiate $x y$).

Now we just need to sort out this mess to get an equation $\frac{\mathrm{dy}}{\mathrm{dx}} = \ldots$

$x \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = - 6 x - 2 - y$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{6 x + 2 + y}{x}$ for all $x \in \mathbb{R}$ except zero.