# How do you differentiate 5x-y=xyln(xy)?

Mar 5, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 - y - y \ln \left(x y\right)}{1 + x + x \ln \left(x y\right)}$

#### Explanation:

Differentiate both sides of the equation with respect to $x$:

$\frac{d}{\mathrm{dx}} \left(5 x - y\right) = \frac{d}{\mathrm{dx}} \left(x y \ln \left(x y\right)\right)$

Use the product rule at the second member:

$5 - y ' = \frac{d}{\mathrm{dx}} \left(x y\right) \ln \left(x y\right) + \left(x y\right) \frac{d}{\mathrm{dx}} \left(\ln \left(x y\right)\right)$

Use the product rule again for $\frac{d}{\mathrm{dx}} \left(x y\right)$ and the chain rule for $\frac{d}{\mathrm{dx}} \left(\ln \left(x y\right)\right)$:

$5 - y ' = \left(y + x y '\right) \ln \left(x y\right) + \left(x y\right) \frac{1}{x y} \frac{d}{\mathrm{dx}} \left(x y\right)$

Simplify:

$5 - y ' = \left(y + x y '\right) \ln \left(x y\right) + \frac{d}{\mathrm{dx}} \left(x y\right)$

$5 - y ' = \left(y + x y '\right) \ln \left(x y\right) + \left(y + x y '\right)$

$5 - y ' = \left(y + x y '\right) \left(1 + \ln \left(x y\right)\right)$

Solve now for $y '$:

$5 - y ' = y \left(1 + \ln \left(x y\right)\right) + x y ' \left(1 + \ln \left(x y\right)\right)$

$5 - y \left(1 + \ln \left(x y\right)\right) = x y ' \left(1 + \ln \left(x y\right)\right) + y '$

$5 - y \left(1 + \ln \left(x y\right)\right) = y ' \left(x + x \ln \left(x y\right) + 1\right)$

$y ' = \frac{5 - y - y \ln \left(x y\right)}{x + x \ln \left(x y\right) + 1}$