Because square root of negative number is complex number, expression under the square root must be non-negative.
`4-x^2>=0`` ` (1)
This is the same as asking where is parabola `y=4-x^2` above x-axis. It is easy to see that null-points are `x=pm2` so (1) holds for `x in [-2,2]`.
So domain of function `f` is `D_f=[-2,2]`. Also derivative of function `f` is
Now to find where `f` is differentiable we need to find domain of `f'`. Its domain are all `x` such that `4-x^2>0` (notice we now have `>` instead of `>=` that is because denominator cannot be 0 i.e. we cannot divide by 0). So domain of `f'` is `D_(f')=(-2,2)`.
So `f` is differentiable on `D_f cap D_(f')=[-2,2]cap(-2,2)=(-2,2)`