You can either use the inverse function theorem or compute the general derivative using implicit differentiation. The first method is slightly faster.
The IFT goes like this: if f(x) is invertible and f(a) = b, then finv(b) = a (where "finv" means "inverse of f").
By definition of inverse functions, we have
f(finv(x)) = finv(f(x)) = x
Differentiating both sides of the second equality with respect to x using the chain rule gives
finv'(f(x)) * f'(x) = 1
When x = a, we get
finv'(b) * f'(a) = 1
or
finv'(b) = 1/f'(a)
Now let f(x) = sin(x), which is invertible over the interval -π/2 ≤ x ≤ π/2. In the interval, we have sin(x) = √3/2 when x = π/3. We also have f'(x) = cos(x).
So we take a = π/3 and b = √3/2. Then
arcsin'(√3/2) = 1/cos(π/3) = 1/(1/2) = 2
