Derivatives in actual calculus don’t have to be continuous either. Consider the function defined by f(x) = x^2 sin(1/x) for x != 0; f(0) = 0.

The derivative at 0 exists and is 0, because lim h-> 0 (h^2 sin(1/h))/h = lim h-> 0 (h sin(1/h)), which equals 0 because the sin function is bounded.

When x !=0, the derivative is given by the product and chain rules as 2x sin(1/x) - cos(1/x), which obviously approaches no limit as x-> 0, and so the derivative exists but is discontinuous.