# What is the implicit derivative of 4= xyln(xy) ?

Nov 18, 2016

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

#### Explanation:

$4 = x y \ln \left(x y\right)$
$\therefore 4 = x y \left(\ln x + \ln y\right)$

Applying the triple product rule we get:

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