# Why is there a dx/dx in the Implicit Differentiation rule?

1. Mar 1, 2012

### DrummingAtom

I tried deriving this one on my own and I'm just not understanding where the dx/dx term comes from. I'm looking dy/dx.

Starting with F(x,y) = 0:

$\frac{\partial{F}}{\partial{x}}\frac{dx}{dx} + \frac{\partial{F}}{\partial{y}}\frac{dy}{dx} = 0$

It seems redundant to say dx/dx when it turns out to be one anyway. Why does this step need to be done? Thanks

2. Mar 1, 2012

3. Mar 1, 2012

### HallsofIvy

It doesn't need to be done. What text did you see that in?

4. Mar 1, 2012

### DrummingAtom

Briggs Calculus. It just seemed kinda strange to say that x is a function of x so we need to take the derivative of it.

5. Mar 1, 2012

i do it!