Why are there two different ops for normal and partial derivatives?

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Treadstone 71
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This may seem like an odd question, but why are there two different ops for the normal and partial derivatives? i.e., [tex]\frac{d}{dx}[/tex] and [tex]\frac{\partial}{\partial x}[/tex]? I don't see a difference if only one is used, since we are always differentiating wrt a single variable anyway.
 
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think of it...

in classical mechanics, you can have functions like the action [itex]L[/itex] which is defined in the form

[tex]L=L(x,y,z,t)[/tex],

but [itex]x=x(t)[/itex], [itex]y=y(t)[/itex], [itex]z=z(t)[/itex], so

[tex]\frac{d L}{d t}=\frac {\partial L}{\partial x} \dot{x}(t)+\frac {\partial L}{\partial y} \dot{y}(t)+\frac {\partial L}{\partial z} \dot{z}(t)[/tex]

which clearly is different from [itex]\partial L/\partial t[/itex].


EDIT:

Sorry, my mistake... the derivative is missing one term. It should be read

[tex]\frac{d L}{d t}=\frac {\partial L}{\partial x} \dot{x}(t)+\frac {\partial L}{\partial y} \dot{y}(t)+\frac {\partial L}{\partial z} \dot{z}(t)+\frac{\partial L}{\partial t}[/tex]
 
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Interesting. I never encountered those. Then again, I'm not in physics.
 
The above are functions present in Hamiltonian systems, which are a big subject of study for mathematitians too... Specially in P.D.E.

EDIT:

Not to mention Calculus of Variations.
 
one came first, d/dx, and the other is a generalization of it, but asking what d/dx of some object is is strictly different from asking what partial d by dx of it is since the former assumes that the other variables (if there are any) are a function of x too. That is to say that if f(x)=x+y then

[tex]\frac{\partial f}{\partial x}[/tex]

makes sense but

[tex]\frac{df}{dx}[/tex]

doesn't
 
Wait, if [tex]f(x)=x+y[/tex], wouldn't [tex]\frac{df}{dx}[/tex] make sense since y is a constant? That is, [tex]\frac{df}{dx}[/tex] does not make sense if it was [tex]f(x,y)=x+y[/tex]?
 
Here's another reason for the disticntion.

I give you y, just y, now differentiate it with respect to x. What's the answer? dy/dx or 0?

I suppose it is unfair of me to expect you to recognize silly errors from catastrpohically bad ones, not to mention hypocritical perhaps.
 
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