Finding $\frac{dy}{dx}$ using implicit differentiation

Question

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 $\displaystyle{\frac{dy}{dx}}$ to one side

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

3. Isolate $\displaystyle{\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?

Answer 1

Use the product rule: $$ \frac{d}{dx}(2xy) = 2x\frac{dy}{dx} + 2\cdot1\cdot y. $$

Answer 2

When you take derivative you will get

$$2xy'+2y+2yy'=1+y'$$

Be careful about $y^2\to2yy'$ and $2xy\to2y+2xy'$