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Partial derivative of function w.r.t. the percent change of the variable

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Rewrite this in terms of f, f, ∂f/∂x, and x:
    ∂f(x,y)/∂(%Δx) = ∂f(x,y)/∂(d log(x) )

    2. Relevant equations

    ∂(%Δf(x,y))/∂(%Δx) = ∂logf(x,y)/∂log(x)= ∂f(x,y)/∂x*x/f(x,y).

    ∂f(x,y)/∂log(x)=x∂f(x,y)/∂x

    3. The attempt at a solution

    I found that (%Δx) can be written as the differential of log(x):
    ∂f(x,y)/∂(%Δx) = ∂f(x,y)/∂(d log(x) ).

    But the partial of derivative of differential of the variable throws me off. It seems like the valid problems that people may have in real world: how much the level of function changes if you increase 1% of the variable, but I can't find anything on it.
     
  2. jcsd
  3. Jan 25, 2012 #2
    I made a mistake in the relevant questions for one of the equations:

    %Δf(x,y)/%Δx = ∂logf(x,y)/∂log(x)= ∂f(x,y)/∂x*x/f(x,y)
    should be corrent.
     
  4. Jan 25, 2012 #3
    I am now confused. The quantity that I want to get is that
    how much f(x,y) changes if you increase x by 1%.
    If you say, "how much f(x,y) changes if you increase x by 1 unit", then
    it should be ∂f(x,y)/∂x.
    So, what I want is: Δf(x,y)/%Δx.
    Can you write this as, ∂f(x,y)/∂log(x)? Then, the answer is x∂f(x,y)/∂x.
     
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