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Inverse Trig Differentiation

  1. Oct 21, 2009 #1
    Alright, I'm not technically stuck on this one, but I consistently get a result that disagrees with what Wolfram Alpha shows when I enter the problem to check my answer. Sorry 'bout the lack of LaTeX, but it should be simple enough to read. Here goes:

    Problem:
    Differentiate y=sin-1[x/(1+x)]

    Basically, I rearrange for
    sin(y)=x/(x+1)

    then use implicit differentiation to yield:
    * ---> cos(y)*(dy/(dx))=1/(x+1)2

    Substituting with:
    cos(y)=sqrt[1-sin2(y)]

    I get:
    cos(y)=sqrt[1-x2/(x+1)2]

    which simplifies to:
    cos(y)=sqrt(2x+1)/(x+1)

    Dividing both sides of the original equation (above, marked with a star) by cos(y):
    dy/(dx)=1/[(x+1)sqrt(2x+1)]

    Which, if you don't like surds in the denominator, can be simplified to:
    sqrt(2x+1)/[(x+1)(2x+1)]


    I've done this question several times, and re-checked all my working. For the life of me, I can't see where I go wrong, yet my result is slightly different to what it should be. Any suggestions would be most welcome.
     
  2. jcsd
  3. Oct 21, 2009 #2
    This is pretty close (and equivalent) to the first alternate form of the derivative on Wolframalpha.
     
  4. Oct 21, 2009 #3

    Dick

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    What do you think it should be?
     
  5. Oct 21, 2009 #4
    Huh. It appears you're right. I had even tried entering <my answer> - <w.a. answer>, and it came out nonzero, but I must have mistyped something. Well, I'm glad I'm not missing anything. Thanks for the clarification!
     
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