Find dy/dx in the following:

Question

y=[x/√(1-x²)]- sin‾¹x

y'={[(1-x²)^(1/2)-(-2x)(x)(1/2)(1-x²)^(-1/2)]/(1-x²)} - (1/√(1-x²))

Then I multiplied (-2x)(x)(1/2) together. How do I simplify the rest?

Answer 1

I agree so far with
y' = {[(1-x²)^(1/2)-(-2x)(x)(1/2)(1-x²)^(-1/2)]/(1-x²)} - (1/√(1-x²))
= {[(1-x²)^(1/2)-(-x^2(1-x²)^(-1/2)]/(1-x²)} - (1-x^2)^(1/2)/(1-x²)
I formed a common denominator for the last term

= (1-x^2)^(-1/2) [1-x^2 - x^2 - (1-x^2)]
= -(1+x^2)/(1-x^2)^(3/2)

There would be other variations, the trick is to recognize if they are the same

Answer 2

Thank you for the help!