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)

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