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July 23, 2014

July 23, 2014

Posted by **sh** on Thursday, April 8, 2010 at 10:23pm.

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?

- Calculus -
**Reiny**, Thursday, April 8, 2010 at 10:40pmI 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

- Calculus -
**sh**, Thursday, April 8, 2010 at 10:57pmThank you for the help!

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