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

DrummingAtom
Messages
657
Reaction score
2
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
 
Physics news on Phys.org
DrummingAtom said:
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
It doesn't need to be done. What text did you see that in?
 
HallsofIvy said:
It doesn't need to be done. What text did you see that in?

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.
 
HallsofIvy said:
It doesn't need to be done.

i do it! o:)
 

Similar threads

Back
Top