# How do you implicitly differentiate xy^2- yln(x-y)^2= 4-x?

Jan 7, 2016

Remember to use the chain rule and product rule when differentiating the y variable ...

#### Explanation:

$x {y}^{2} - y \ln {\left(x - y\right)}^{2} = 4 - x$

Now, differentiate ...

${y}^{2} + x \left(2 y\right) y ' - \ln {\left(x - y\right)}^{2} - \frac{y}{x - y} ^ 2 \left(2\right) \left(x - y\right) \left(1 - y '\right) = - 1$

Next simplify and solve for $y '$ ...

$y ' \left[\left(2 x y\right) + \frac{2 y}{x - y}\right] = \ln {\left(x - y\right)}^{2} + \frac{2 y}{x - y} - 1$

$y ' = \frac{\ln {\left(x - y\right)}^{2} + \frac{2 y}{x - y} - 1}{\left(2 x y\right) + \frac{2 y}{x - y}}$

hope that helped