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

  • Thread starter Usuiisu
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  • #1
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Homework Statement



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

Homework 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

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.
 

Answers and Replies

  • #2
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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.
 
  • #3
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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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