Solving Partial Derivatives with f(x-z)=x+y+z

imana41
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hi i have a problem for this
if f(x-z)=x+y+z solve
gif.latex?\frac{\partial%20z}{\partial%20x}-2\frac{\partial%20z}{\partial%20y}.gif


can i say u=x-z and write F(x,y,z)=x+y+z-f(u) and then
%20z}{\partial%20x}=-\frac{\frac{\partial%20F}{\partial%20x}}{\frac{\partial%20F}{\partial%20z}}.gif

%20z}{\partial%20y}=-\frac{\frac{\partial%20F}{\partial%20y}}{\frac{\partial%20F}{\partial%20z}}.gif


or this isn't true ? thanks if u help me.
 
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f(x-z)=x+y+z, this doesn't make sense, the way you've written it f is a function of one variable (x-z), but has 3 variables and does not reference the original variable (x-z)
 
what should i do ?
 
can you start by trying to write the question exactly as it was written?
 
For example, are you given that f(x, y, z)= x+ y+ z= 0?
 
HallsofIvy said:
For example, are you given that f(x, y, z)= x+ y+ z= 0?

no jus written f(x-z)=x+y+z and f is a function of x and y(x,y are Independent)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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