calculez : f"

si : f(sinx)=(sinx)²
calculez : f"

f(t)=t^2 qqs t de [-1,1] ==> f''=2

If F(sin(x)) = sin(x²) in [-π/2 ,π/2] then
F(x) = sin(arcsin²(x)) in [0 , 1]. What about F''?

In ]-1 ,1[ : dF/dx = 2arcsin(x) (1 /cos[arcsin(x)]) cos[arcsin²(x)] = 2arcsin(x) *cos[arcsin²(x)] /cos[arcsin(x)], then in ]-1 ,1[ :
½ cos²[arcsin(x)].d²F/dx² = [1 /cos[arcsin(x)] *cos[arcsin²(x)] +arcsin(x) *2arcsin(x) /cos[arcsin(x)] *(-sin[arcsin²(x)])] cos[arcsin(x)] -arcsin(x) *cos[arcsin²(x)] *(-sin[arcsin(x)] /cos[arcsin(x)]) = cos[arcsin²(x)] -2arcsin²(x) *sin[arcsin²(x)] +x *arcsin(x) *cos[arcsin²(x)] /cos[arcsin(x)] = cos[arcsin²(x)] -2arcsin²(x) *sin[arcsin²(x)] +x *arcsin(x) *cos[arcsin²(x)] /√(1 -x²) since cos[arcsin(x)] >= 0, therefore = √(1 -sin²[arcsin(x)]) for all x, finally
d²F/dx² = 2[ cos[arcsin²(x)] -2arcsin²(x) *sin[arcsin²(x)] +x *arcsin(x) *cos[arcsin²(x)] /√(1 -x²) ] /(1 -x²) in ]-1 ,1[