Deriving Inverse Trig Functions

[quozzy]

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.

 

[Bohrok]

dy/(dx)=1/[(x+1)sqrt(2x+1)]

This is pretty close (and equivalent) to the first alternate form of the derivative on Wolframalpha.

 

[Dick]

What do you think it should be?

 

[quozzy]

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!