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December 21, 2014

December 21, 2014

Posted by **Cupcake** on Tuesday, April 1, 2008 at 1:00pm.

I would appreciate any help that can be offered.

- Calculus 1 - implicit differentiation -
**Reiny**, Tuesday, April 1, 2008 at 1:31pmif you can do the basic rules like the product rule, then you can do implicit differentiating.

xy+2x+3x^2=4

Let me explain term by term

xy is a product, so when you differentiate

you get

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

or

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

2x would give you 2 and

3x^2 would give you 6x (really it was 6x(dx/dx) which is 6x(1) or 6x)

and of course the derivative of 4 is zero

so you would have

x(dy/dx) + y + 2 + 6x = 0

solving this for dy/dx you would finally have

dy/dx = (-6x - y - 2)/x

You will get questions where there is a dy/dx in several terms.

In that case bring all those terms to one side of your equation, factor out the dy/dx, and solve for dy/dx that way.

the key this is to remember:

if you differentiate an x term you also get a dx/dx hanging around, which is 1, so you don't have to write it, but

if you differentiate a y term you would get a dy/dx hanging around.

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