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Nth derivative of ln(x+c)

  1. Dec 15, 2008 #1
    Is there a formula or a method to derive the nth derivative of ln (x+c)...(where c is any number...such as 1, 2, 3, etc)?

    I know that there is a formula for the nth derivative of ln x, but I was wondering if this formula can be extended to one we can use when we are dealing with the nth derivative of something slightly more complicated.
     
  2. jcsd
  3. Dec 15, 2008 #2
    Use the chain rule or the definition of the derivative to show that f(n)(x + c) is the same as f(n)(u) if you let u = x + c.
     
    Last edited: Dec 15, 2008
  4. Dec 15, 2008 #3
    You should try to derive a formula for the derivative of y = ln(g(x)). The way I would do that is using implicit differentiation, namely:

    [tex] y = ln(g(x)) [/tex]

    [tex] e^{y} = g(x) [/tex]

    [tex] \frac{dy}{dx} e^{y} = \frac{dg}{dx} [/tex]

    [tex] \frac{dy}{dx} = \frac{g'(x)}{g(x)} [/tex]

    Hope this helps.
     
  5. Dec 15, 2008 #4

    lurflurf

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    yes
    [log(x+C)]'=1/(x+C)
    [log(x+C)]''=-1/(x+C)^2
    (D^n)[log(x+C)]=(n-1)!(-1)^(n+1)/(x+C)^n
    for n=1,2,3,4,...

    is can be done for other simple functions like sin(x), exp(x), and x^a. The trouble with more complicated functions is complicated sums arise.
     
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