Put `u = 1 - x^2`

Then we have: `y = sin^-1 u`

`y=sin^-1(1-x^2)`

`(dy)/(dx)=1/sqrt(1-u^2)(du)/(dx)`

`=1/sqrt(1-(1-x^2)^2)(-2x)`

`=(-2x)/(sqrt(1-(1-x^2)^2)`

Expanding the expression under the square root and simplifying gives:

`(dy)/(dx)=(-2x)/(sqrt(-x^4+2x^2)`

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