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Derivatives of ln including Absolute Value

  1. Nov 16, 2011 #1
    1. The problem statement, all variables and given/known data
    Determine the domain and find the derivative

    f(x) = ln|(x+2)/(x3 - 1)|


    2. Relevant equations



    3. The attempt at a solution

    We can factor x3 - 1 = (x-1)(x2) + x + 1)
    From this we know that x ≠ 1 or -2 because 1 would be undefined at -2 would cause the function to be ln 0 which is not possible.
    Domain = {x[itex]\inℝ[/itex], x ≠ 1,-2}
    f(x) = ln(x+2) - ln(x3 - 1)
    = ln(x+2) - ln[(x-1)(x2) + x + 1)]
    = ln(x+2) - ln(x-1) - ln(x2) + x + 1)
    f'(x) = 1/(x+2) - 1/(x-1) -(2x+1)/(x2) + x + 1)

    This is where i stopped but i know that there is an absolute value so i'm not sure if i have to find when x is positive and when x is negative and then find the seperate derivatives. Please help me figure out if up to here is fine or if i have to keep going and if i do have to keep going, how would i go about finding the derivative? Thank you.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 16, 2011 #2
    Nope, no separate derivatives needed. I don't remember the proof, but if its a natural log involving an absolute value, the derivative would be the same for f(x) if abs() wasn't there.

    Do you want me to find it?

    Also.. why are you factoring things?

    Just use the chain rule.

    d/dx (ln|w|) = 1/w * dw/du * du/dx

    feel free to consult http://www.wolframalpha.com/input/?i=d/dx+(f(x)+=+ln|(x+2)/(x3+-+1)|)

    if you need further help...
     
    Last edited: Nov 16, 2011
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