Using chain rule, what is the derivative of sin(arcsin(x))

Chain rule
y=f(g(x))
y´=(d f(gx)/d g)(d g/d x)

or

y=y(v)   and v=v(x), then dy /dx=(dy/dv)(dv/dx)

in our case:
y=sin (v)
v=arcsin(x)

dy/dv=d sin (v)/dv=cos (v)=cos(arcsin(x)
dv/dx=d arcsin(x)/dx=1/√(1-x²)

dy/dx=[cos (arcsin(x))]/√(1-x²)

d sin(arcsin(x))/dx=[cos (arcsin(x))]/√(1-x²)

What can we do about it

The answer would be the second choice : )