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Inverse Trig Differentiation

  • Thread starter quozzy
  • Start date
  • #1
15
0
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.
 

Answers and Replies

  • #2
867
0
dy/(dx)=1/[(x+1)sqrt(2x+1)]
This is pretty close (and equivalent) to the first alternate form of the derivative on Wolframalpha.
 
  • #3
Dick
Science Advisor
Homework Helper
26,258
618
What do you think it should be?
 
  • #4
15
0
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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