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Partial derivative chain rule for gradient

  1. Jan 6, 2013 #1
    1. The problem statement, all variables and given/known data

    compute the gradient:

    [tex] ln(z / (sqrt(x^2-y^2)) [/tex]




    2. Relevant equations

    [tex]∇=(∂/(∂x)) + ... for y and z [/tex]

    I just want to know how to do the first term with respect to x

    3. The attempt at a solution

    I am so rusty I dont know where to begin.
     
  2. jcsd
  3. Jan 6, 2013 #2
    [itex]ln(z) - \frac{1}{2}ln(x^2-y^2)[/itex]

    That ought to make it easier. Now treat y and z as constants.
     
  4. Jan 7, 2013 #3
    You don’t sum... Gradient is a vector and what you have found are the linearly independent components of it.
     
  5. Jan 7, 2013 #4

    Fredrik

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    Staff Emeritus
    Science Advisor
    Gold Member

    jfgobin, you shouldn't give away the final answer when you're answering posts in the homework forum. Just give a hint, like e^(i Pi)+1=0 did.
    Mod note: I dealt with this.
    physics2000, The hint given by e^(i Pi)+1=0 simplifies the problem significantly, but you don't have to use it. It's also possible to use these three rules directly:
    \begin{align}
    &\log'(x)=\frac 1 x\\
    &(fg)'(x)=\frac{f'g-fg'}{g^2}\\
    &\frac{d}{dx}x^a=ax^{a-1}
    \end{align}
     
    Last edited by a moderator: Jan 7, 2013
  6. Jan 7, 2013 #5
    Yup, my bad!
     
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