If someone could review my work on this exercise

When I use f X(x) is the pdf of X and fY(y) is the pdf of Y.

Consider X a random variable with pdf

f X(x) = (3/8) * (1-x)^2 if -1<=x<=1

and 0 otherwise.

Find the pdf of Y = 1-X^2

**My workings:**
If -1<=x<=1 then 0<=Y<=1

Then P(Y<=y) = P(1-X^2<=y) = P(X^2>=1-y) = P(X>=sqrt(1-y))

=1-P(X<=sqrt(1-y))

=1 - (3/8) * (1 - sqrt(1-y)) ^ 2 for 0<=y<=1

and 0 otherwise

Now f Y (y) = d/dy (P(Y<=y))

= (3/4) * (1-sqrt(1-y)) / (2*sqrt(1-y)) for 0<=y<=1 and 0 otherwise!

**Or should I say P(X^2>=1-y) = 1-P(X^2<=1-y) = 1- P(-sqrt(1-y) <=X<= +sqrt(1-y)) and then how do I proceed?**