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Derivative w.r.t. a function 
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#1
Aug207, 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! 


#2
Aug307, 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 . 


#3
Aug307, 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!! 


#4
Aug307, 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. 


#5
Aug307, 03:21 PM

P: 12

i have realized that what i posed was not meaningful. i am now thinking over my problem again.. thanks!



#6
Aug307, 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



#7
Aug307, 04:22 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,568

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] 


#8
Aug307, 05:34 PM

P: 1,705




#9
Aug507, 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) 


#10
Aug507, 07:55 PM

P: 173




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