A-Math Diff Problem

• Jt061952

  8 Jul 09, 19:10

  I am having difficulty differentiating this problem.  Can someone kindly help?

  Differentiate:     ln ( x + sqrt (x^2 - 1))

  The answer given in the book is given as: 1 / sqrt (x^2  - 1).   Thank you.

   

• Forbiddensinner

  8 Jul 09, 19:26

  Edited.

   

   

• wee_ws

  8 Jul 09, 19:34

  Hi Jt061952,

    Let y = ln [x + sqrt(x^2 - 1)].

    Then dy/dx = [ 1 + (2x)/sqrt(x^2 - 1) ] / [ x + sqrt(x^2 - 1) ], which simplifies to your said answer.

    In general, dy/dx = f'(x) / f(x) if y = ln f(x). Thanks!

  Cheers,
  Wen Shih 

• Forbiddensinner

  8 Jul 09, 20:02

  Edited : Drats. I am 100% sure now I made a mistake somewhere.

  My apologies to Lecturor Wee.

  On the otherhand, do note that it is

  [ 1 + x/sqrt(x^2 - 1) ] / [ x + sqrt(x^2 - 1) ]

  rather than

  [ 1 + (2x)/sqrt(x^2 - 1) ] / [ x + sqrt(x^2 - 1) ]

• Jt061952

  8 Jul 09, 21:17

  Thanks, I plugged the equation into Wolfram but I did not get the simplified answer. 

• wee_ws

  8 Jul 09, 21:18

  Hi,

    You may wish to check with Wolfram Integrator at:

    http://integrals.wolfram.com/index.jsp

    Wolfram indeed returns the answer ln [x + sqrt(x^2 - 1)].

    Thanks!

  Cheers,
  Wen Shih

• Jt061952

  8 Jul 09, 21:24

  WS,

     Initially I tried using the chain rule as in Wolfram and it got horrendous and the answer did not come out in the simplified form.  But will check again.  Thanks again.

• wee_ws

  8 Jul 09, 21:33

  Hi,

    Just type in 1/(Sqrt[x^2 - 1]) and the answer comes out in an instant.

    Anyhow, my working above is correct. Thanks!

  Cheers,
  Wen Shih

• Jt061952

  8 Jul 09, 21:42

  WS,

        Got it!  Thanks a bunch Sifu.  Good nite.  jt