Calculus
posted by sh on .
Find dy/dx in the following:
y=[x/√(1x²)] sin‾¹x
y'={[(1x²)^(1/2)(2x)(x)(1/2)(1x²)^(1/2)]/(1x²)}  (1/√(1x²))
Then I multiplied (2x)(x)(1/2) together. How do I simplify the rest?

I agree so far with
y' = {[(1x²)^(1/2)(2x)(x)(1/2)(1x²)^(1/2)]/(1x²)}  (1/√(1x²))
= {[(1x²)^(1/2)(x^2(1x²)^(1/2)]/(1x²)}  (1x^2)^(1/2)/(1x²)
I formed a common denominator for the last term
= (1x^2)^(1/2) [1x^2  x^2  (1x^2)]
= (1+x^2)/(1x^2)^(3/2)
There would be other variations, the trick is to recognize if they are the same 
Thank you for the help!