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Calculus/implicit differenciation

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Expert: - 7/19/2009


Question
Hey, I'm currently taking an AP calculus course and I've encountered some problems when it comes to implicit differenciation, so here is my prob:

while knowing that dy/dx will be added to one of the variables when the equation is differenciated for the first time, like:
If x^2 - 2xy + 3y^2 = 2, then
2x - 2(x * dy/dx + y*1) + 6y * dy/dx = 0.

Are there rules that state when to put the dy/dx symbol? I know that dy/dx can not be distributed to every variables when differenciated, but how can I know WHICH variable WILL receive it?

thanks

Answer
When faced with problems like this one, get rid of the parenthesis first.  Next, move all terms with no dy/dx in them to the other side.
On the new left side, factor out dy/dx from all the terms, putting the parenthesis back in.  Then divide by whatever the dy/dx is being multiplied by.  To illurate how to do this, I will use the problem above.

Multiply the function out (so that the parenthesis or dropped).

This gives 2x - 2x(dy/dx) - 2y + 6y(dy/dx) = 0.

Moving the terms without dy/dx to the right side gives
-2x(dy/dx) + 6y(dy/dx) = -2x + 2y.

Dividing all terms by -2 gives x(dx/dy) - 3y(dy/dx) = x - y.

Factoring out the left side gives (x - 3y)(dy/dx) = x - y.

Dividing both sides by (x-3y) gives dy/dx = (x-y)/)x-3y).

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