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Derivatives of natural logarithmic functions

  1. Sep 8, 2008 #1
    f(x) = ln (12x-5/9x-2)

    f'(x) = (4/3) (1/ln10)(9x-2/12x-5)

    Is this correct??
  2. jcsd
  3. Sep 8, 2008 #2


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    Instead of ln(10) you should put ln(base of the logarithm), in this case; ln(e) = 1. When deriving log(x) (base 10), you can rewrite to ln(x)/ln(10). ln(10) is just a constant, so the derivative of this is 1/ln(10) * 1/x.

    Also, I think you didn't do the chain rule quite right.
  4. Sep 8, 2008 #3
    Then, is this
    f(x) = ln (12x-5)
    g(x) = ln (9x-2)
    f'(x) = 12/12x-5
    g'(x) = 9/9x-2

  5. Sep 8, 2008 #4
    Is this correct:

    Last edited: Sep 8, 2008
  6. Sep 8, 2008 #5


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    This should not have ln inside. What is d/dx ln(12x-5) ? You've got it from above, just put it into this one.
  7. Sep 9, 2008 #6


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    I have the feeling you are guessing more than working systematically. Actually there are two ways to solve this. One is by using the chain rule:
    set y = (12x - 5) / (9x - 2). Then the derivative of ln(y) is 1/y . dy/dx.

    The other way is to first use ln(a/b) = ln(a) - ln(b) and then use the sum rule to differentiate (you still need the chain rule!).
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