• Try show that for the statement to be true, the derivative has to be defined.
• There are 2 definitions for the derivative, either would be useful in the proof

https://gmtex.siri.sh/fs/1/School/Extra/Maths/Qotd solutions/2024-05-15_differentiablility-implies-continuity.html

Here’s a pretty minimum effort hand-wavy proof.

A function is differentiable on an interval if it’s differentiable at every point in that interval. Similarly, a function is continuous on an interval if it’s continuous at every point in that interval. So I’ll show that differentiability at a point implies continuity at that point, from which differentiability on an interval implying continuity on an interval will automatically follow.

A function is differentiable at x = a if lim (x → a) (f(x) - f(a)) / (x - a) exists, so let us assume it does exist.

In that limit, the denominator equals 0 at the limit point. Yet, we know by assumption that the limit exists. This implies the numerator also equals 0 at the limit point, because the only way for our denominator to equal 0 at the limit point, and still have the limit exist, is for the expression to be indeterminate, rather than undefined. The only indeterminate form with 0 in the denominator is 0/0.

So we can assume lim (x → a) (f(x) - f(a) = 0. As f(a) is a constant value, this implies lim (x → a) f(x) = f(a) - so the function is continuous at x = a.