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Derivative of ln(u)^k?

  1. Oct 4, 2012 #1
    This is a concept that confuses me. If you have a function where f(x) = ln(u)^k (where u is a function and k is an unknown constant), then how would you solve for the derivative? I've already guessed it would be

    f'(x) = [kln(u)^(k-1)](1/u)(u')

    Is that right?
     
  2. jcsd
  3. Oct 5, 2012 #2
    looks right to me
     
  4. Oct 5, 2012 #3

    Mark44

    Staff: Mentor

    No, it's not. The derivative does not involve a log.

    If f(x) = ln[uk], where u is a differentiable function of x, then
    $$f'(x) = \frac{1}{u^k}\cdot ku^{k-1}\cdot u'$$
    $$= \frac{k}{u}\cdot u'$$

    This is essentially the same problem that you asked in the other thread you posted. As I said there, you can do things this way, but it is much simpler to simplify the log expression before you differentiate.

    If we simplify first, we have
    f(x) = k ln(u)
    So f'(x) = k * d/dx(ln(u)) = k * 1/u * u' = (k/u)* u', which is the same as I got by differentiating without simplifying first.

    For readers who didn't see the other thread, I'm assuming that what you wrote as ln(u)^k means ln[uk], and not [ln(u)]k.
     
  5. Oct 6, 2012 #4
    Ha, yes, I misread the OP - I thought the exponent was on the lnu. My mistake.
     
  6. Oct 6, 2012 #5
    I think the power is on the ln(u), not on u itself. He closed the parenthesis before adding the power, so it appears to be :
    f(x) =(ln(u))^k.
     
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