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

Apr 26, 2015

You have to use the implicit function theorem (aka the Dini's theorem)

The implicit equation is $F \left(x , y\right) = 1 - x - x y + y = 0$, F is a smooth function (it's a polynomial), so we can use Dini's theorem

Notation: ${F}_{x} = \frac{\mathrm{dF}}{\mathrm{dx}}$

Dini's theorem says that in every point $\left(x , y\right)$ such that ${F}_{y} \ne 0$, we have a neighbourhood where $y = f \left(x\right)$ with $f$ smooth and $f ' = - {F}_{x} / {F}_{y}$

So, ${F}_{y} = - x + 1 \implies \setminus \forall \left(x , y\right) \ne \left(1 , y\right)$
$f ' \left(x\right) = - \frac{- y - 1}{- x + 1} = \frac{1 + f \left(x\right)}{1 - x}$

Now we invert, ${F}_{x} = - y - 1 \implies \forall \left(x , y\right) \ne \left(x , - 1\right)$
$x = g \left(y\right) \mathmr{and} g ' \left(y\right) = - \frac{1 - x}{- y - 1} = \frac{1 - g \left(y\right)}{1 + y}$

So in $\left(1 , - 1\right)$ the function is not differentiable, elsewhere is differentiable (it's ${C}^{1}$).

Notice that we don't have an "explicit" formula for the derivatives

Apr 26, 2015

$1 - x y = x - y$

When we differentiate, we'll need the product rule for the second term on the left.
I use the order: $\left(f g\right) ' = f ' g + f g '$ for this term we'll have $f = - x$ and $g = y$

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

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

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

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

So

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

And

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