# How do you implicitly differentiate 1=-xy+x+y?

##### 1 Answer
Mar 17, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1 - y}{1 - x}$

#### Explanation:

In implicit differentiation, one variable is not linked to another explicitly and it may not look like say $y = f \left(x\right)$ but it will appear more as say $F \left(x , y\right) = 0$.

The question is then how we evaluate $\frac{\mathrm{dy}}{\mathrm{dx}}$. We do this by differentiating both sides w.r.t $x$ and wherever $y$ appears, its differential is put as $\frac{\mathrm{dy}}{\mathrm{dx}}$ and then evaluate $\frac{\mathrm{dy}}{\mathrm{dx}}$, which may be function of both $x$ and $y$.

Let us do this using the given function $1 = - x y + x + y$. Differentiating both sides we get

$0 = - \left(\frac{\mathrm{dx}}{\mathrm{dx}} \cdot y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) + 1 + \frac{\mathrm{dy}}{\mathrm{dx}}$ (here we use product formula on $- x y$. So, we have

$0 = - y - x \frac{\mathrm{dy}}{\mathrm{dx}} + 1 + \frac{\mathrm{dy}}{\mathrm{dx}}$ or

$0 = \left(1 - x\right) \frac{\mathrm{dy}}{\mathrm{dx}} + 1 - y$

or $- \left(1 - x\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1 - y$ and hence

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1 - y}{1 - x}$