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BucketOfFish
BucketOfFish is offline
#5
Mar3-12, 09:27 PM
P: 50
Thanks for the response, but I feel like maybe I didn't express my question well. Let me try again, using the chain rule definition you provided.

[tex]\frac{df}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}[/tex]

Here, when y is not a function of x (the two are independent), the partial and total derivatives are equivalent.

[tex]\frac{df}{dx}=\frac{\partial f}{\partial x}[/tex]

However, when y [itex]is[/itex] a function of x, then the inconsistency I mentioned above comes into play. [itex]\frac{\partial f}{\partial x}[/itex] does not have a well-defined value, but depends upon how f is expressed. (It does not make sense to me how you can vary x while holding y constant, if y is a function of x.) Thus, if the value of a partial derivative is either non-consistent, or else equivalent to a total derivative, then I don't see what the point of taking a partial derivative is, outside of defining a total derivative (and even then the matter seems fishy to me).

EDIT: Never mind, I thought about taking a gradient in non-Cartesian coordinates and realized that partial derivatives are in fact helpful. I still hold that they have no use outside of re-expressing a total derivative.