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Derivatives of ln(x)

  1. Jan 21, 2009 #1
    1. The problem statement, all variables and given/known data
    I know how to take the derivative of ln(x), it's just 1/x. But what if you had something more complicated than just x?

    For example, ln(x4(2x+5)5)?


    2. Relevant equations



    3. The attempt at a solution

    I guess you would still do 1/(x4(2x+5)5), then multiply it by the derivative of the denominator.

    Which would be 4x3(2x+5)5 + x4(5(2x+5)4)(2).

    Is that correct?

    The problem I'm supposed to be doing is actually more complicated, it's ln[x5(x+4)3(x3+4)6]. Would the procedure be similar? I guess I'm not sure about taking the derivative of something with 3 terms, I've only ever seen it done with two.
     
  2. jcsd
  3. Jan 21, 2009 #2
    Yes. You are correct:

    For example:

    y= ln(x2)

    y' = (1 / x2) .dx2dx

    y' = 2x/x2

    Also with three terms:

    y= u.v.w

    y' = u'.v.w + u.v'.w + u.v.w'

    or you can deal with u and v and then w.
    I think the u'.v.w + u.v'.w + u.v.w' will be a bit quicker

    your question: ln[x5(x+4)3(x3+4)6]

    let u = x5 , v= (x+4)3 , w = (x3+4)6

    (Note that any of these could be u, v w) But remember not to forget the natural log function when differentiating :)
     
    Last edited: Jan 21, 2009
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