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Differentiation of Ln(x+1/x)

  1. Nov 2, 2008 #1
    Hi all,

    I have to differentiate

    [tex]Ln(x+\frac{1}{x})[/tex]

    where I first differentiate Ln and than multiply by the differentiation of the inner function

    [tex]1/(x+\frac{1}{x})*(1-\frac{1}{x^2}[/tex]

    which I simplify to

    [tex]\frac{x}{x^2+1}*(1-\frac{1}{x^2})[/tex]

    [tex]\frac{x}{x^2+1}-\frac{1}{x*(x^2+1)}[/tex]

    the problem is I cannot rewrite it to this

    [tex]\frac{2*x}{x^2+1}-\frac{1}{x}[/tex]

    how to rewrite it - any help or advise appreciated. Thanks in advance

    Best
    Jorgen
     
    Last edited: Nov 2, 2008
  2. jcsd
  3. Nov 2, 2008 #2
    Start at the final expression, and write it as a fraction with one denominator: x(x^2+1)
     
  4. Nov 2, 2008 #3
    thanks,

    so I put into one fraction

    [tex]\frac{x^2-1}{x*(x^2+1)}[/tex]

    but I don't know how to start rearranging this.... Any new hints

    Best

    J
     
  5. Nov 2, 2008 #4

    gabbagabbahey

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    [itex]x^2-1=(x^2+1)-2[/itex] ;0)
     
  6. Nov 2, 2008 #5
    so I rewrite the fraction using this hint

    [tex]\frac{(x^2+1)-2}{x*(x^2+1)}[/tex]

    I split the fraction into

    [tex]\frac{1}{x}-\frac{2}{x*(x^2+1)}[/tex]

    but I can still not see how to rearrange it

    Thanks in advance

    Best
    J
     
  7. Nov 2, 2008 #6

    gabbagabbahey

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    Use partial fractions to decompose your second term; in other words, find the constants [itex]A[/itex],[itex]B[/itex], and [itex]C[/itex] that satisfy:

    [tex]\frac{2}{x*(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}[/tex]
     
  8. Nov 3, 2008 #7
    In what form do you want it? Whether it is condensed into one fraction or written as a difference doesn't matter if both expressions are equal.
     
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