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What is the difference between a partial differental and an ordinary differential?

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Pinu7
#1
Jun8-09, 09:42 PM
P: 270
I have been wanting to ask this for a while.

In Calc I, I was introduced to differentials. It seemed like they act like quantities(please corrected me if I'm wrong). For example dx/dx=1. You can obtain this by differentiating x or by eliminating the dx in the numerator and denominator(I do not know why this worked).

What convinced me that differentials where quantities was the chain rule. dy/dx=(dy/du)(du/dx). The proof is a bit tough, but you will obtain the same result by eliminating the du.(I may be making a TREMENDOUS mathematical blunder here, but it seemes to work)

In Calc III, I was introduced to [tex]\partial[/tex]x and[tex]\partial[/tex]y. Obviously I found out that [tex]\partial[/tex]x[tex]\neq[/tex]dx or else the chain rule for multiple variables would not simplify to dz/du.

So, why are these two infinitesimals so different?
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HallsofIvy
#2
Jun9-09, 05:55 AM
Math
Emeritus
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Thanks
PF Gold
P: 39,552
No, you were not "introduced to [itex]\partial x[/itex] and [itex]\partial y[/itex] in Calc III. You were introduced to the partial derivatives [itex]\partial f/\partial x[/itex] and [itex]\partial f/\partial y[/itex]. There is no such thing as a "[itex]\partial x[/itex]".

One important reason is that the partial derivatives themselves just don't tell you enough about the function. If the derivative of a function of one variable exists at a point, then it is differentiable (and so continuous) at that point. A function of several variables can have all its partial derivatives at a point and still not be differentiable nor even continuous at that point.

Take f(x,y)= 0 if xy= 0, 1 otherwise. It is easy to show that [itex]\partial f/\partial x= \partial f/\partial y= 0[/itex] at (0,0) but f is not even continuous there.
Pinu7
#3
Jun9-09, 05:35 PM
P: 270
Thanks, that cleared things up for me, HallsofIvy.


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