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Partial Derivative Rules

  1. Jun 5, 2012 #1
    I am having trouble finding the rule for the (partial) derivative of an expression like

    y = f(X)^{g(X)}

    can anyone help?
     
  2. jcsd
  3. Jun 5, 2012 #2

    micromass

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    Hint:

    [tex]f(x)^{g(x)}=e^{f(x)g(x)}[/tex]
     
  4. Jun 5, 2012 #3

    Mark44

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    Make that eg(x) * ln(f(x)) and I'll agree.
     
  5. Jun 5, 2012 #4

    micromass

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    Oh my, I should pay better attention while posting :blushing:
     
  6. Jun 5, 2012 #5
    hmm

    I was just looking at:
    wolfram alpha for
    [tex]
    d/dx \ f(x,y)^{g(x,y)} = f^{(1,0)}(x,y)g(x,y)f(x,y)^{g(x,y)-1} + g^{(1,0)}(x,y)f(x,y)^{g(x,y)}log(f(x,y) \\
    d/dy \ f(x,y)^{g(x,y) } = f^{(0,1)}(x,y)g(x,y)f(x,y)^{g(x,y)-1} + g^{(0,1)}(x,y)f(x,y)^{g(x,y)}log(f(x,y) \\
    d/dx \ e^{f(x,y)g(x,y)} = f^{(1,0)}(x,y)g(x,y)e^{f(x,y)g(x,y)} + f(x,y)g^{(1,0)}(x,y)e^{f(x,y)g(x,y)}
    [/tex]

    and was unsure what the [tex] d/dx \ f(x,y) \ is\ f^{(1,0)or(0,1)}(x,y)[/tex] meant in the previous equations.
    Is it just the partial derivative? or partial at the point (1,0) or (0,1)?
     
  7. Jun 5, 2012 #6
    my intuition is that it's a bit vector to show which variable the derivative is with respect to
     
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