MATH. DY/DX HELP!

• dylanninetytwo

  28 May 10, 13:19

  dy/dx in(inx^e) .

   

  help appreciated

• wee_ws

  28 May 10, 13:39

  Hi,

    Do you mean: find dy/dx if y = ln (ln x^e)? Thanks.

  Cheers,
  Wen Shih

• dylanninetytwo

  28 May 10, 13:45

  Nope. Just differeniate in(inx^e).

   

  the ans is 1/ x In x.

   

  thanks alot!

• wee_ws

  28 May 10, 13:54

  Hi,

    I believe it's to differentiate ln (ln x). Natural logarithm is written as 'ln'. Thanks.

  Cheers,
  Wen Shih

• dylanninetytwo

  28 May 10, 13:58

  but the ans isnt 1/x Inx

• wee_ws

  28 May 10, 14:03

  Hi,

    It's clearer if you write the answer as (1/x) (ln x).

    In that case, the question may have been to differentiate (1/2) (ln x)^2.

    Recall that d/dx [ (ln x)^n ] = n . (ln x)^(n - 1) . (1/x).

    Thanks.

  Cheers,
  Wen Shih

• dylanninetytwo

  28 May 10, 14:14

  Thanks !

• SBS261P

  28 May 10, 14:42

  de original question was correct.

  d [ln (ln x^e)] / dx

  = d [ ln (e ln x) ] / dx

  = d [ ln e + ln (ln x) ] / dx

  = d [1 + ln (ln x) ] / dx

  = (1/ln x ) . 1/x

  = 1 / (x ln x)

  does this look correct? (",)...

• dylanninetytwo

  29 May 10, 09:31

  That is the correct method. Thanks alot!!!! YEAH!