Inverse of a partial derivative?

[romsofia]

As we know, the inverse of a derivative is an integral and visa versa, but what's the inverse of a partial derivative? Is it even possible to un-do a partial derivative?

Thanks for your help as I've been thinking about this for a couple days now!

 

[zhermes]

The difference between a partial and a complete derivative is what the derivative is with respect to. This is the same as for an integral---i.e. the reverse of a complete derivative would be an integral over all variables, while the reverse of a partial derivative would be an integral over only the one variable in question.

 

[Mute]

To add on to what zhermes said, an integral will undo a partial derivative, but the arbitrary constant is now an arbitrary function. e.g., given \partial f(x,y,z)/\partial x,

\int dx~\frac{\partial f(x,y,z)}{\partial x} = f(x,y,z) + C(y,z)

 

[romsofia]

The difference between a partial and a complete derivative is what the derivative is with respect to. This is the same as for an integral---i.e. the reverse of a complete derivative would be an integral over all variables, while the reverse of a partial derivative would be an integral over only the one variable in question.

To add on to what zhermes said, an integral will undo a partial derivative, but the arbitrary constant is now an arbitrary function. e.g., given \partial f(x,y,z)/\partial x,

\int dx~\frac{\partial f(x,y,z)}{\partial x} = f(x,y,z) + C(y,z)

Thanks, another question, why is it a function?

 

[Mute]

Thanks, another question, why is it a function?

In single-variable calculus, you have to add an arbitrary constant when you differentiate because you know the derivative of a constant is zero, so when doing indefinite integration you have to account for the possibility that in undoing the derivative there was a constant that was lost because it was differentiated.

The same thing applies in multivariable calculus: if I have a function C(y,z), then the partial derivative of C(y,z) with respect to x is zero:

\frac{\partial C(y,z)}{\partial x} = 0

so when I integrated \partial f(x,y,z)/\partial x with respect to x to undo the derivative I had to account for the fact that there may have been a function of y and z that was lost when the full function was differentiated.

 

[HallsofIvy]

Thanks, another question, why is it a function?

If you are referring to the "C(y,z)" it is because partial differentiation with respect to one variable treats other variables like constants:

x^2y+ e^zcos(y)
x^2y+ ln(yz)+ sin(z^2)
x^2y- cos(e^z)+ y^2
all have partial derivative, with respect to x, 2xy.

 

[romsofia]

Is this something usually taught in calc three (the how to "un-do" a partial derivative)? I can't seem to find any of this in my notes :x