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Math Help - Derivative involving ln

  1. #1
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    Derivative involving ln

    I need help with this problem.

    find the derivative of f(u) = ln(u)/1 + ln(2u)


    I started by finding ln of ln(u) and 1 + ln(2u) . . .

    I got ln(u) = 1/u and 1 + ln(2u) = 1. Then I just used the quotient rule, but I can't get the right answer. The answer is supposed to be 1 + ln2/ u[1 + ln(2u)]^2.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by zachb View Post
    I need help with this problem.

    find the derivative of f(u) = ln(u)/1 + ln(2u)


    I started by finding ln of ln(u) and 1 + ln(2u) . . .

    I got ln(u) = 1/u and 1 + ln(2u) = 1. Then I just used the quotient rule, but I can't get the right answer. The answer is supposed to be 1 + ln2/ u[1 + ln(2u)]^2.

    Using the quotient rule, which is

    d/dx f(x)/g(x) = [g(x)f '(x) - f(x)g'(x)]/g(x)^2

    f(u) = ln(u)/1 + ln(2u)
    => f '(u) = {[1 + ln(2u)](1/u) - ln(u)[1/u]}/(1 + ln(2u))^2
    => f '(u) = {(1 + ln(2u) - ln(u))/u}/(1 + ln(u))^2
    => f '(u) = [1 + ln(2u) - ln(u)]/u(1 + ln(u))^2
    => f '(u) = [1 + ln2]/u(1 + ln(u))^2 ........................we got to this step by combining the ln's. remember, lnx - lny = ln(x/y)


    where did you make the mistake? first off, the derivative of ln(2u) is 1/u by the chain rule. if this was not your fatal mistake, perhaps you used the quotient rule incorrectly
    Last edited by Jhevon; March 17th 2007 at 05:43 PM.
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  3. #3
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    firstly you write down the derivatives of ln(u) and 1 + ln(2u);

    d/du ln(u) = 1/u
    d/du 1+ ln(2u) = d/du 1+ ln(u) + ln(2) ;by using the rule log[ab]=log[a]+log[b]
    = 1/u

    then quotient rule you get;
    [1/u[1+ln(2u)] - (ln(u))(1/u) ]/(1+ln2u)^2

    factorise;
    [1/u(1+ln(2u)-ln(u) )]/(1+ln2u)^2

    again use a log rule; ln[a]-ln[b] = ln[a/b] and put the 1/u as a u down the bottom;
    [1+ln(2u/u)]/[u(1+ln2u)^2]
    =[1+ln(2)]/[u(1+ln2u)^2]

    which is your result.
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