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