Differentiating Ln

  • Thread starter Focus
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  • #1
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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
 

Answers and Replies

  • #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)
 
  • #3
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SebastianG said:
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)
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:
 
  • #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]
 
  • #5
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HallsofIvy said:
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]
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)
 
  • #6
NateTG
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Focus said:
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)

It seems like it pops right out from the chain rule.
 
  • #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!
 
  • #8
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Hmmm...I tried that but I'll do it again, cheers for the help :smile:
 

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