Problem from Zorich's book volume 1 Show that if a function $f$ is defined and differentiable on an open interval $I$ and $[a,b]\subset I$, then
a) the function $f'(x)$ (even if it is not continuous!) assumes on $[a,b]$ all the values between $f'(a)$ and $f'(b)$;
b) if $f''(x)$ also exists in $(a,b)$, then there is a point $\xi\in (a,b)$ such that $f'(b)-f'(a)=f''(\xi)(b-a).$
My approach:
a) I know that this is the well-known Darboux's theorem and I was able to prove it.
b) But I have issues with this part. It looks very similar to mean value theorem but MVT is not applicable here because $f'(x)$ is continuous on the open interval $(a,b)$.
I guess since it comes after Darboux's theorem then probably it could derived via part a) but I cannot see this.
Would be thankful for the solution.
 A: For part $(b)$:  Suppose on the contrary that no such $\xi$ exists then for all $t\in (a,b)$, we have 
$f''(t)\ne \frac{f'(b)-f'(a)}{b-a}$
If for some $t_1,t_2\in (a,b)$ such that $t_1\ne t_2$ we have $f''(t_1)\gt \frac{f'(b)-f'(a)}{b-a}\gt f''(t_2)$ then by Darboux theorem there exists some $x_0\in (t_1,t_2)$ such that $f''(x_0)=\frac{f'(b)-f'(a)}{b-a}$ (call it $p$ for brevity), which violates our assumption. So WLOG, let $f''(x)\gt p$ for $x\in (a,b).$ 
Define: $g(x)=f'(x)-p(x-a)\implies g'(x)\gt 0$ for all $x\in (a,b)$ 
$g(a)=g(b)=f'(a)$ and noting that $g$ is strictly increasing on $(a,b)$, now $\exists c \in (a,b): g(c)\ne g(a)$. (If not then $g(x)=g(a)$ for all $x\in (a,b)$)
WLOG, let $g(c)\gt g(a)=g(b),$ (the other case is similar) by Darboux theorem $\exists c_1\in (c,b)$ such that $g(c)\gt g(c_1)\gt g(b)$ which violates that $g$ is strictly increasing on $(a,b)$.
Therefore, it is not true that $f''(x)\gt p$ for all $x\in (a,b)$. 
Similarly, it is not true that $f''(x)\lt p$ for all $x\in (a,b)$. Hence by contradiction $\exists \theta \in (a,b)$ such that $f''(\theta)=p$

