# Finding dy/dx using implicit differentiation

I'm having trouble getting the correct answer and any help would be appreciated.

Given a function: $2xy + y^2 = x + y$

Here are my steps:

1) Find the derivative of each value

$$2 + 2\frac{dy}{dx} + 2y\frac{dy}{dx} - 1 - 1\frac{dy}{dx} = 0$$

2) move $\frac{dy}{dx}$ to one side

$$1 = \frac{dy}{dx} (-2 -2y + 1)$$

3) isolate $\frac{dy}{dx}$

$$\frac{1}{-2 -2y + 1} = \frac{dy}{dx}$$

But the correct answer is: $$\frac{1 - 2y}{2x + 2y - 1}$$

Now obviously with simplifying the answer will differ a little, but I cannot seem where I went wrong.

used the product rule twice to find $2xy$'s derivative and the rest is simple exponential rules.

Thoughts?

Use the product rule: $$\frac{d}{dx}(2xy) = 2x\frac{dy}{dx} + 2\cdot1\cdot y.$$
$$2xy'+2y+2yy'=1+y'$$
Be careful about $y^2\to2yy'$ and $2xy\to2y+2xy'$
• since $y$ depends on $x$, $f(y)'=f'(y).y'$ that is why $y^2\to 2yy'$ – mesel Oct 5 '14 at 22:25
• $$(2xy)'=(2x)'y+2x(y)'=2y+2xy'$$ – mesel Oct 5 '14 at 22:30