Assume $f$ has a continuous second derivative $f~''$ in some neighborhood of $a$.Then, for every $x$ in this neighborhood, we have $f(x) = f(a) + f~'(a)(x-a) + E_1(x)$ , where
$$E_1(x) = \int_a^x (x-t)f~''(t)dt$$
Derivation: $E_1(x) = f(x)-f(a) - f~'(a)(x-a) = \int_a^x f~'(t)dt - f~'(a)\int_a^x dt = \int_a^x[f'(t)-f'(a)]dt$
Using integration by parts ( take $u = f~'(t)-f~'(a) , dv =dt= d(t-x)$, we get :
$E_1(x) = \int_a^x (x-t)f~''(t)dt$.
I understand why $f~''$ must be continuous owing to the fundamental theorem of calculus applicable to the expression : $E_1(x) = \int_a^x (x-t)f~''(t)dt$
However, shouldn't $f~'(x)$ be also continuous as we have expressed :
$f(x)-f(a) - f~'(a)(x-a) = \int_a^x f~'(t)dt - f~'(a)\int_a^x dt = \int_a^x[f'(t)-f'(a)]dt$
By the fundamental theorem of calculus, must not $f~'(t)$ be also continuous as $f(x)-f(a) = \int_a^x f~'(t)dt?$
Thank you for reading through!
In general differentiability $\implies$ continuity.
So if the second derivative $f''$ exists, then $f'$ is continuos