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Homework Help: Differentiating Ln

  1. Oct 25, 2005 #1
    I've had troubled sleep because of this...:cry:
    ln.jpg
    I tryed a lot and got this...
    metryln.jpg
    Can you spot any mistakes or give me hints on how to aproach this:yuck:
    Thanks
     
  2. jcsd
  3. Oct 25, 2005 #2
    That is an equation that shows the "complex variable" version of arcsinh (x) as equal to the integral version of it (on real values). Like a definition. You obtain the first with the complex functions sinhz and some algebra, and the second one with a trigonometric substitution (or using mathematica, hehe)
     
  4. Oct 25, 2005 #3
    I am so confused...I know that the second one can be obtained by using y=arsinh(x) but I cant get the log of the function to differentiate to that:uhh:
     
  5. Oct 25, 2005 #4

    HallsofIvy

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    It helps a lot if you write it clearly! I think you mean that IF
    [tex]y= ln\left[x+ (1+x^2)^{\frac{1}{2}}\right][/tex]
    then
    [tex](1+x^2)\left(\frac{dy}{dx}\right)^2= 1[/tex]
    which is the same as saying that
    [tex]\frac{dy}{dx}= \frac{1}{(1+x^2)^{\frac{1}{2}}}[/tex]
     
  6. Oct 25, 2005 #5
    The problem is I can't derive [tex]\frac{dy}{dx}= \frac{1}{(1+x^2)^{\frac{1}{2}}}[/tex] from [tex]y= ln\left[x+ (1+x^2)^{\frac{1}{2}}\right][/tex]...
    I can only derive it from y=arsinh(x)
     
  7. Oct 25, 2005 #6

    NateTG

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    It seems like it pops right out from the chain rule.
     
  8. Oct 25, 2005 #7

    HallsofIvy

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    Yep.
    [tex]y= ln\left[x+ (1+x^2)^{\frac{1}{2}}\right][/tex]so
    [tex]y'= \frac{1}{x+(1+x^2)^{\frac{1}{2}}}\left(1+(\frac{1}{2})(1+ x^2)^{-\frac{1}{2}}(2x)\right)\)[/tex]
    Now factor that
    [tex](1+ x^2)^{-\frac{1}{2}}[/tex]
    out of the numerator and every thing else cancels!
     
  9. Oct 25, 2005 #8
    Hmmm...I tried that but I'll do it again, cheers for the help :smile:
     
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