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Differentiating [ sin(1/ln(x)) / x ] solution?

  1. Jul 24, 2005 #1
    [SOLVED] differentiating [ sin(1/ln(x)) / x ].. solution?

    hello all. I do not have the solution to this question that im about to ask. But if you find the time, try this solving this problem and feel free to type your answer and compare with mine.
    differentiate :: sin(1/ln(x)) / x

    my answer: -[ ( (sin(1/lnx)*(ln(x)^2) + cos(1/ln(x)) ) / ( (x^2)*(ln(x)^2) ) ]

    i am extremely curious about what you all got. I want to see if i did it correctly. Thanks in advance.

    I found some time to find the d^2y/dx^2 of this f().

    answer #2 for d^2y/dx^2 : { [ (cos(1/lnx))(2+4lnx+2(lnx)^2+(x^2)(lnx)^3) ]+[ (sin(1/lnx))* (lnx) * (4(lnx)^2 - x^2+2(lnx)^3) ] } / { [(lnx)*x]^3 }

    woah, i spent so much time doing it.. if you guys have a time to do it, post your answer pls! thanks again.
     
    Last edited by a moderator: Jul 24, 2005
  2. jcsd
  3. Jul 24, 2005 #2

    HallsofIvy

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    Pretty straight forward, just a little tedius. (ln x)'= 1/x so (1/ln x)'= ((ln x)-1)'= -(ln x)-2(1/x)= -1/(x(ln x)2).

    (sin(1/ln x))'= cos(1/ln x)(1/ln x)'= -cos(1/ln x)/(x(ln x)2).


    By the quotient rule, (sin(1/ln x)/x)'= ((sin(1/ln x))'x- sin(1/ln x))/x2=
    -cos(1/ln x)/(x3(ln x)2)- sin(1/ln x)/x2.
    I think, except for the fact that I separted the two fractions, that's the same thing you have.

    By the way, this belongs in the "Calculus" section, not the "Differential Equations" section. The fact that the problem involves a derivative does not mean it is a differential equation!
     
  4. Jul 24, 2005 #3
    sorry about that, if you don't mind, could you move this thread to the "Calculus" section? Thanks.
     
  5. Jul 24, 2005 #4
    Besides your expression for the derivative don't forget to state [tex]x > 0 , x \neq 1[/tex] as a derivative at [tex]x_{1}[/tex] needs to be defined over some interval [tex]x_{1}-R<x<x_{1}+R[/tex]
     
  6. Jul 24, 2005 #5
    [tex]\boxed{\frac{d}{dx} \left( \frac{ \sin \left[ \frac{1}{\ln x} \right]}{x} \right) = - \frac{1}{x^2} \left( \frac{ \cos \left[ \frac{1}{\ln x} \right]}{(\ln x)^2} + \sin \left[ \frac{1}{\ln x} \right] \right)}[/tex]
     
  7. Jul 24, 2005 #6
    First derivative is OK.
    But the second isn't.
    I got
    [tex]\frac{d^2}{dx^2} \left( \frac{ \sin \left[ \frac{1}{\ln x} \right]}{x} \right) = \frac{ 2\cos( \frac{1}{\ln x}) - \frac{\sin( \frac{1}{\ln x})}{\ln x}+ 2(\ln x)^3 \sin ( \frac{1}{\ln x}) +3 \ln x\cos \left[ \frac{1}{\ln x} \right]) }{x^3 (\ln x)^3} [/tex]
    mathematica gives the same
     
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