# How do you implicitly differentiate 1=-xy+x+y-y^2+x^2?

##### 1 Answer
Dec 5, 2015

$y ' = \frac{y - 2 x - 1}{- 2 y - x + 1}$

#### Explanation:

The process of implicit differentiation involves taking the derivative, $D$, of both sides of the equation at the same time, then solving for $y '$.

$D \left\{1\right\} = D \left\{- x y + x + y - {y}^{2} + {x}^{2}\right\}$

The left hand side is a constant, so the derivative is zero. The right hand side can be split up into the sum of several derivatives.

$0 = D \left\{- x y\right\} + D \left\{x\right\} + D \left\{y\right\} - D \left\{{y}^{2}\right\} + D \left\{{x}^{2}\right\}$

Now we go through and take the derivative of each term one by one. We need to use the product rule on the first term. The second and fifth term can be solved using power rule, and the chain rule can be used to solve the fourth term. The third term is by definition $y '$.

$0 = - y - x y ' + 1 + y ' - 2 y y ' + 2 x$

Now group all of the terms that have a $y '$ in them and pull out the $y '$.

$0 = - x y ' + y ' - 2 y y ' - y + 1 + 2 x$

$0 = y ' \left(- x + 1 - 2 y\right) - \left(y - 1 - 2 x\right)$

We can move the two grouped terms to either side of the $=$.

$y ' \left(- x + 1 - 2 y\right) = \left(y - 1 - 2 x\right)$

Divide both sides by $\left(- x + 1 - 2 y\right)$. I rearranged inside the parenthesis to make the answer look neater.

$y ' = \frac{y - 2 x - 1}{- 2 y - x + 1}$