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Derivative w.r.t. a function

by ledol83
Tags: derivative, function
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ledol83
#1
Aug2-07, 10:47 PM
P: 12
Hi, i have a question on taking derivative w.r.t. to a function (instead of an independent variable). Actually i saw an excellent post on this same forum but that one was about a single variable.

My question is: f is function of x and y, and z is some other function also dependent on x and y, so is the following correct?

df/dz=df/dx*dx/dz+df/dy*dy/dz

it differs from the classic chain rule in the sense that z is actually a function (not an independent var), so i am not sure about this.

I appreciate so much for any comment!
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lalbatros
#2
Aug3-07, 01:49 AM
P: 1,235
First, go back to how the derivative wrt to a function is defined. (functional derivative)
Second, be more precise about your specific question.
As I understood, f is a function of x and y: f(x,y)
therefore, if you confirm that, I would say the df/dz = 0 .
ledol83
#3
Aug3-07, 09:16 AM
P: 12
Hi actually i have f(x,y) and z(x,y) and was just wondering if this is true:

df/dz=df/dx*dx/dz+df/dy*dy/dz

thanks a lot!!

lalbatros
#4
Aug3-07, 03:09 PM
P: 1,235
Derivative w.r.t. a function

No this cannot be true, since this has no meaning.
Tell us what you think the meaning of df/dz would be, maybe then we can help.
ledol83
#5
Aug3-07, 03:21 PM
P: 12
i have realized that what i posed was not meaningful. i am now thinking over my problem again.. thanks!
ice109
#6
Aug3-07, 03:23 PM
P: 1,705
how could you have a derivative of a function with respect to another function that the first function is not a function of
HallsofIvy
#7
Aug3-07, 04:22 PM
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Thanks
PF Gold
P: 39,363
If f(x) is a function of x and g(x) is a function of x, you can surely write f as a function of g.

In particular,
[tex]\frac{df}{dg}= \frac{df}{dx}\frac{dx}{dg}= \frac{\frac{df}{dx}}{\frac{dg}{dx}}[/tex]
ice109
#8
Aug3-07, 05:34 PM
P: 1,705
Quote Quote by HallsofIvy View Post
If f(x) is a function of x and g(x) is a function of x, you can surely write f as a function of g.

In particular,
[tex]\frac{df}{dg}= \frac{df}{dx}\frac{dx}{dg}= \frac{\frac{df}{dx}}{\frac{dg}{dx}}[/tex]
not that i'm doubting you personally but i don't see where this comes from; proof?
lalbatros
#9
Aug5-07, 02:55 PM
P: 1,235
"not that i'm doubting you personally but i don't see where this comes from; proof?"

The first point is to know when we are talking about f1=f(g) or f2=f(x),
The second point is about the class of functions considered,
Otherwise, this is trivial (assuming dg is smooth):

f'(g) = f(g+dg)/dg = f(g(x)+dg(x))/dg(x) = f(g(x) + g'(x) dx) / (g'(x) dx) = f'(x)/g'(x)
AiRAVATA
#10
Aug5-07, 07:55 PM
P: 173
Quote Quote by ice109 View Post
not that i'm doubting you personally but i don't see where this comes from; proof?
It is a consequence of the chain rule and the inverse function theorem. Maybe you should do some reading too instead of making fun of peoples questions. Better try to help or keep out.


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