Regular derivative vs. partial derivative

[DocZaius]

Through my learning of calculus, I have come under the impression that there is an important difference between the derivative of a variable with respect to another, and the partial derivative of a variable with respect to another. For example:

I think that \frac{dy^2}{dx} = 2y\frac{dy}{dx}

and that \frac{\partial y^2}{\partial x} = 0

Is that correct? And if it is, then why do practically all online derivative calculator say that: \frac{dy^2}{dx} = 0?

Note: They actually use the regular derivative symbols, not the partial derivative symbols. If they did the latter, they would be correct.

 

[Amanheis]

You are correct.

The answer to the second question is, because they assume dy/dx = 0. In Mathematica for example, you'd have to write y[x] instead of just y to tell Mathematica that y is a function of x and you should get the result you expect.

 

[DocZaius]

But isn't that dy/dx = 0 assumption incorrect? If we don't know whether y is a function of x or not, should we not take precautions in case it is, and come up with 2y*dy/dx ?

If y isn't a function of x, that still comes out to 0. But if y is a function of x, then 2y*dy/dx is correct and 0 is wrong!

Very troubling assumptions by those online calculators.

 

[lurflurf]

From a particular point of view total derivative and partial derivatives are the same. The dependencies of the variables and the nature of derivatives should be clearly stated, but it is often they are not. One may make the convention that when the total notation is used, variables are dependent unless otherwise stated; and when partial notation is used they are independent unless otherwise stated.
The dy/dx = 0 assumption is not incorrect. The user that does not adust for it is incorrect.
If a calculator given
3*4+3
returns
15
but the user intended
3*(4+3)
which returns 21
the user was in error.

 

[DocZaius]

...and when partial notation is used they are independent unless otherwise stated.

I wasn't even aware that it was possible to treat variables as dependent with partial notation. Could you give the simplest example possible of partial notation being used with variables being treated as dependent?

The dy/dx = 0 assumption is not incorrect. The user that does not adust for it is incorrect.
If a calculator given
3*4+3
returns
15
but the user intended
3*(4+3)
which returns 21
the user was in error.

The order of operations convention isn't a very good comparison to make, since the answer the calculator returns there is either entirely correct or entirely wrong in regards to the user's intention (and that intention misplaced if he disagrees with order of operations, a very strong convention)

My proposition is that dy^2/dx simply return 2y*dy/dx which is always correct! Whether one wishes to make an assumption (and it is an assumption) that y is independent of x, should be one's own decision to make, not the calculator's.

I would have also made the point that we actually have the notation and convention for deriving a variable independently with relation to another, and that that is the partial derivative. And that as a result, and because of the fact that y's independence of x in regular derivatives is rightly in question, the regular derivative calculation should at the very least give the safe, always correct answer. However, you seem to imply that partial derivatives can treat variables dependently. If that's possible, I take this part back. :)

 

[HallsofIvy]

Through my learning of calculus, I have come under the impression that there is an important difference between the derivative of a variable with respect to another, and the partial derivative of a variable with respect to another. For example:

I think that \frac{dy^2}{dx} = 2y\frac{dy}{dx}

and that \frac{\partial y^2}{\partial x} = 0

Is that correct? And if it is, then why do practically all online derivative calculator say that: \frac{dy^2}{dx} = 0?

Note: They actually use the regular derivative symbols, not the partial derivative symbols. If they did the latter, they would be correct.

This has nothing to do with the distinction between "ordinary" and "partial" derivatives. If you assume that y is a function of the single variable x, then d(y^2)/dx= 2y dy/dx by the chain rule. If y is NOT a function of x, then dy/dx= 0 and so d(y^2)/dx= 0.

If you assume that y is a function of x (and some other variables), then \partial (y^2)/\partial x= 2y \partial y/\partial x. But if you assume that f(x,y)= y² where x and y are independent variables, then \partial y/\partial x= 0 and so \partial (y^2)/\partial x= 0 just as with the ordinary derivative.

Again, it is not a distinction between "ordinary" and "partial" derivatives, it is a matter of whether you are assuming y is a function of x or assuming that x and y are independent variables.