Use implicit differentiation to find dy/dx

2xy - y^2 = 1

2xy - 2y * y' = 1 ?

How do I do this problem?

Question

Use implicit differentiation to find dy/dx 

2xy - y^2 = 1

2xy - 2y * y' = 1 ?

How do I do this problem?

terenzreignz · Answer

The gist of it is... you differentiate x like normal and you differentiate y as you would for x, except after you're done differentiating the y-part, you multiply it to dy/dx or y'
For instance, the derivative of
x^3 is still 3x^2
but the derivative of
y^3 is (3y^2)(dy/dx)
It's like the chain rule, see..

anonymous · Answer

okay
well so far I have 2y + 2xy' - 2y * y' = 1

Not sure if that's right.

terenzreignz · Answer

You forgot to differentiate the right-side of the equation.  Otherwise, you've done well :P

anonymous · Answer

What would that be? I was going to leave the one there and move my x's and y's to the right.

terenzreignz · Answer

If it were
2xy - y^2 = x, 
then certainly, differentiating gives
2y + 2xy' - (2y)(y') = 1
But it's 1 on the other side, and you differentiated the left...

terenzreignz · Answer

In general, when you have two functions... such that
f(x) = g(x)
You can take the derivative of one side, but are you SURE it's still equal to the other function?
Most of the time it isn't, BUT it's true that
f'(x) = g'(x)

raden · Answer

derivative of 1 is 0