`f(x)=sqrt(4-x^2)`

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

`f'(x)=1/(2sqrt(4-x^2))cdot(-2x)=-x/(sqrt(4-x^2))`

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)`

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