Hello, dadon!

That arcsin expression is very messy to differentiate, so I modified it.Differentiate: .f(x) .= .x·(1 - x²)^½ - arcsin(1 - x²)^½

We have: .θ .= .arcsin(1 - x²)^½ . → . sin θ .= .(1 - x²)^½

θ is an angle is a right triangle with: .opp = (1 - x²)^½ .and .hyp = 1.

. . Pythagorus tell us that: .adj = x

Hence: .cos θ = x . → . θ = arccos(x)

The function becomes: . f(x) .= .x·(1 - x²)^½ - arccos(x)

Then: . f'(x) .= .x·½(1 - x²)^{-½)(-2x) + (1 - x²)^½ + (1 - x²)^{-½)

Rearrange: .(1 - x²)^{-½} - x²·(1 - x²)^{-½} + (1 - x²)^½

Factor: . (1 - x²)·(1 - x²)^{-½} + (1 - x²)^½

. . . . = .(1 - x²)^½ + (1 - x²)^½

. . . . = .2(1 - x²)^½