Understanding Partial Differentials of a Function

[drag]

Greetings !

I'd appreciate some help in explaining, in general,
what - extracting partial differentials of a function, means.

I'm talking about a function like f(x,y,z).
Does it mean that I need a single solution where
I will have differentials for x,y,z of the func.?
Example:
f'(x,y,z) = ( x^2*y + y^2*z + x*e^(2z) )' =
= (2x*y + e^(2z))x' + (x^2 + 2y*z)y' + (y^2 + 2e^(2z))z'

Also (this is related to physics), if I have unit vectors
of x, y, z in the func. do they stay as they were
or does it entail doing some tricks on them as well
(for Cartesian coordinates - I don't think I should touch'em,
but what about a func. of polar coordinates - discribing the
course of the particle itself).

Thanks !

Live long and prosper.

 

[selfAdjoint]

There will be three first partials, with respect to x, to y, and to z. Each one of them will be a function (in general) of all three variables. You do the x partial by differentiating as if y and z were just constants. So the partial of x^2(lny)sinz is 2x(lny)sinz, for example, but the partial of that with respect to z is x^2(lny)cosz. Just apply the rules you know, and hold the variables you're not differentiating with respect to constant.

 

[drag]

Thanks selfAdjoint !

So can I also write a new whole function like I did
in my example or do I have to treat them separately ?

 

[HallsofIvy]

The problem with something like

"f'(x,y,z) = ( x^2*y + y^2*z + x*e^(2z) )' =
= (2x*y + e^(2z))x' + (x^2 + 2y*z)y' + (y^2 + 2e^(2z))z'"

is that then notation is ambiguous. If you are given that x, y, and z are functions of some other variable, say t, then we could write f as a function of t and differentiate that. In that case f' refers to differentiation with respect to t, as do x', y', and z'. That's the chain rule for functions of several variable.

You titled this partial "differentials". You should say "partial derivatives". Differentials are something else entirely.

Given f as you have it, the correct formulation, in terms of differentials, would be:
df(x,y,z) = d( x^2*y + y^2*z + x*e^(2z) ) =
= (2x*y + e^(2z))dx + (x^2 + 2y*z)dy + (y^2 + 2e^(2z))dz

Of course, for any variable t, you could "divide" by dt to get
df/dt(x,y,z) = d( x^2*y + y^2*z + x*e^(2z) )/dt =
= (2x*y + e^(2z))dx/dt + (x^2 + 2y*z)dy/dt + (y^2 + 2e^(2z))dz/dt.

which is exactly the same as
f'(x,y,z) = ( x^2*y + y^2*z + x*e^(2z) )' =
= (2x*y + e^(2z))x' + (x^2 + 2y*z)y' + (y^2 + 2e^(2z))z'