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

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Discussion Overview

The discussion revolves around the presence of the term dx/dx in the context of implicit differentiation, specifically questioning its necessity and redundancy when deriving dy/dx from the equation F(x,y) = 0. The scope includes theoretical aspects of calculus and differentiation rules.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant expresses confusion about the dx/dx term, questioning its redundancy since it equals one.
  • Another participant suggests that the inclusion of dx/dx may not be necessary, asking for the source of the claim.
  • A participant references a specific textbook, Briggs Calculus, indicating that the notation seems strange to them.
  • One participant states they include dx/dx in their work, implying a personal preference or habit.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the necessity of the dx/dx term; some argue it is redundant while others include it in their calculations.

Contextual Notes

There are unresolved questions regarding the conventions of notation in calculus and the implications of including or excluding certain terms in differentiation.

DrummingAtom
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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:

[itex]\frac{\partial{F}}{\partial{x}}\frac{dx}{dx} + \frac{\partial{F}}{\partial{y}}\frac{dy}{dx} = 0[/itex]

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

[itex]\frac{\partial{F}}{\partial{x}}\frac{dx}{dx} + \frac{\partial{F}}{\partial{y}}\frac{dy}{dx} = 0[/itex]

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:)
 

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