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Partial with respect to ln(f(x))

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data
    I'm working a problem, and I've come to taking the derivative with respect to ln(x):

    [itex]\frac{\partial ln(x^{c})}{\partial ln(p_{x})}[/itex]


    2. Relevant equations

    ln(x[itex]^{c}[/itex])=ln(p[itex]^{2}_{y}[/itex]I)+ln(p[itex]_{x}[/itex]+p[itex]_{y}[/itex])-ln(p[itex]_{x}[/itex])-ln(p[itex]_{y}[/itex])


    3. The attempt at a solution
    I've worked it out, but am not sure how the ln(p[itex]_{x}[/itex]+p[itex]_{y}[/itex]) term would derive with respect to ln(p[itex]_{x}[/itex]). Any help would be great. Thanks!

    [itex]\frac{\partial ln(x^{c})}{\partial ln(p_{x})}[/itex] = [itex]\frac{\partial}{\partial ln(p_{x})}[/itex](ln(p[itex]_{x}[/itex]+p[itex]_{y}[/itex])) - 1
     
  2. jcsd
  3. Oct 12, 2011 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Probably you can use the chain rule here, finding a convenient "intermediate" variable, e.g.

    [tex]
    \frac{\partial \ln(p_x + p_y)}{\partial \ln(p_x)} =
    \frac{\partial \ln(p_x + p_y)}{\partial p_x} \cdot
    \frac{\partial p_x}{\partial \ln(p_x)}
    [/tex]

    looks like I could do it
     
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