# How do you differentiate -y=lnxy-xy?

Jun 27, 2016

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

#### Explanation:

this is a bit easier [administratively] if you use the Implicit Function Theorem...

which says that $\textcolor{red}{y ' = - \frac{{f}_{x}}{{f}_{y}}}$ for function $f \left(x , y\right) = c o n s t$

so from
$- y = \ln x y - x y$

we get
$y + \ln x y - x y \setminus \textcolor{b l u e}{= 0} = f \left(x , y\right)$

and so....
${f}_{x} = \frac{1}{x y} \cdot y - y$
$= \frac{1}{x} - y$
and
${f}_{y} = 1 + \frac{1}{x y} \cdot x - x$
$= 1 + \frac{1}{y} - x$

so using the IFT....
$y ' = - \frac{{f}_{x}}{{f}_{y}}$

$= - \frac{\frac{1}{x} - y}{1 + \frac{1}{y} - x}$

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

there are some pretty simple ways to intuit the Implicit Function Theorem

one is to note that if $f \left(x , y\right) = c$

then the total differential $\mathrm{df} = 0$

and $\mathrm{df} = {f}_{x} \mathrm{dx} + {f}_{y} \mathrm{dy} = 0$

so $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{f}_{x}}{{f}_{y}}$