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# implicit differentiation watch

1. y³ + xy = 4x - 2

I understand that to differentiate y³ I would get 3y² dy/dx but how do I do xy? It's on a practise paper I'm doing and I can't seem to work out where the answer comes from. Thanks x
2. To do d/dx[xy] you have to use product rule. You get:

(1)(y) + (dy/dx)(x) = y + x(dy/dx)

(differentiate the first term, times by the other etc...)
3. think of it as a product of two functions and differentiate is as such.

d(xy)/dx = x(dy/dx) + y
4. Thanks a lot :-) I did know that, I just have my p5 paper tomorrow and I've had 4 modules of exams already this week and I'm too tired to think, but I need to revise! Sorry for asking a silly question!
5. are you on AQA too?
6. y³ + xy = 4x - 2

3y² dy/dx + (y + xdy/dx) = 4
3y² dy/dx + xdy/dx = 4-y
dy/dx(3y²+x) = 4-y
dy/dx = (4-y)/(3y²+x)
7. Yes i am

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