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Homework Help: Simplification problems (I hope)

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data

    1. Differentiate g(x)= ln(x(x^2 - 1)^1/2

    2. Differentiate g(x)= x^2(ln(2x))

    2. Relevant equations

    Law of logs and product rule

    3. The attempt at a solution

    1. I simplified to g'(x) = 1/x + (1/2)(1/(X^2 - 1)(2x) then
    g'(x) = 1/x + 2x/(2(x^2 - 1))
    Answer however is g'(x)= (2x^2-1)/x(x^2-1)

    2. I simplified to g'(x) = 2xln(2x) + x(1/2x)
    Answer however is g'(x) = x + 2xln(2x) Im not sure how to get rid of the 2 in the denominator of the 2nd term

    Tjvelcro
     
  2. jcsd
  3. Jul 11, 2010 #2
    [tex]g'(x) = \frac{1}{x} + \frac{2x}{2(x^2 - 1)}[/tex]
    [tex]g'(x) = \frac{2(x^2-1)}{2x(x^2-1)} + \frac{2x^2}{2x(x^2 - 1)}[/tex]
    [tex]g'(x) = \frac{2(x^2 - 1) + 2x^2}{2x(x^2-1)}[/tex]
    [tex]g'(x) = \frac{2x^2 - 2 + 2x^2}{2x(x^2-1)}[/tex]
    [tex]g'(x) = \frac{4x^2 - 2}{2x(x^2-1)}[/tex]
    [tex]g'(x) = \frac{2x^2 - 1}{x(x^2-1)}[/tex]

    g'(x) = 2xln(2x) + x(1/2x)

    That is wrong. Did you mistakenly type the x in the denominator of that second term? With or without that extra x, the answer is still incorrect. Post what you did.
     
  4. Jul 11, 2010 #3
    So for the second problem

    Differentiate g(x)= x^2(ln(2x))
    I used product rule.

    g'(x) = 2x(ln(2x)) + (x^2)(2/2x)
    g'(x) = 2x(ln(2x)) + (x) This is where I made my mistake, I get confused with y'lnx=1/x and y'[lng(x)] = g'(x)/g(x)

    Thanks

    Tjvelcro
     
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