# How do you implicitly differentiate 18=(y-x)ln(xy)?

$y ' = \frac{\frac{y}{x} - 1 - \ln \left(x y\right)}{\frac{x}{y} - 1 - \ln \left(x y\right)}$

#### Explanation:

Start from the given equation and differentiate both sides of the equation with respect to x

$18 = \left(y - x\right) \ln \left(x y\right)$

$\frac{d}{\mathrm{dx}} \left(18\right) = \frac{d}{\mathrm{dx}} \left(\left(y - x\right) \cdot \ln \left(y x\right)\right)$

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

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

$0 = \left(y - x\right) \cdot \frac{x y ' + y \cdot 1}{x y} + y ' \ln \left(x y\right) - \ln \left(x y\right)$

$0 = \left(y - x\right) \cdot \left(\frac{y '}{y} + \frac{1}{x}\right) + y ' \ln \left(x y\right) - \ln \left(x y\right)$

$0 = y ' + \frac{y}{x} - \frac{x y '}{y} - 1 + y ' \ln \left(x y\right) - \ln \left(x y\right)$

Transpose the terms with y'

$\frac{x y '}{y} - y ' - y ' \ln \left(x y\right) = \frac{y}{x} - 1 - \ln \left(x y\right)$

factor out the y'

$\left(\frac{x}{y} - 1 - \ln \left(x y\right)\right) \cdot y ' = \frac{y}{x} - 1 - \ln \left(x y\right)$

Divide both sides by $\left(\frac{x}{y} - 1 - \ln \left(x y\right)\right)$

$\frac{\cancel{\frac{x}{y} - 1 - \ln \left(x y\right)} \cdot y '}{\cancel{\left(\frac{x}{y} - 1 - \ln \left(x y\right)\right)}} = \frac{\frac{y}{x} - 1 - \ln \left(x y\right)}{\frac{x}{y} - 1 - \ln \left(x y\right)}$

$y ' = \frac{\frac{y}{x} - 1 - \ln \left(x y\right)}{\frac{x}{y} - 1 - \ln \left(x y\right)}$

God bless....I hope the explanation is useful.