Finding dy/dx using implicit differentiation
Matthew Martinez
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?
$\endgroup$2 Answers
$\begingroup$Use the product rule: $$ \frac{d}{dx}(2xy) = 2x\frac{dy}{dx} + 2\cdot1\cdot y. $$
$\endgroup$ $\begingroup$When you take derivative you will get
$$2xy'+2y+2yy'=1+y'$$
Be careful about $y^2\to2yy'$ and $2xy\to2y+2xy'$
$\endgroup$ 4