Take $a=0$, $b=1$,
$$f(x) := \begin{cases} x^2 \cdot \sin x^{-\frac{3}{2}} & x \in (0,1] \\ 0 & x=0 \end{cases}$$
Then $f$ is differentiable and of bounded variation, but $f'$ is unbounded.
Hint To show that $f$ is of bounded variation you can use the following theorem: Let $f: [0,1] \to \mathbb{R}$ differentiable and $f' \in L^1([0,1])$. Then $f$ is of bounded variation and $$\text{Var} \, f = \int_0^1 |f'(t)| \, dt$$
Remark As Pavel M suggested one can also prove that $f$ is of bounded variation by splitting up the interval $[0,1]$ in intervals $[a_n,b_n]$ such that $f$ is monotone on $[a_n,b_n]$. Then one can easily compute the variation of $f$ on the interval $[a_n,b_n]$ and use the fact that the variation on $[0,1]$ is equal to the sum of the variations on $[a_n,b_n]$.
We have $\vert f\vert \le M<\infty$ on $[a,b].$
Let $\epsilon >0$ and choose a partition $P = \left \{ a = x_0 < x_1 < \cdots < x_n = b \right \}$ such that
$\tag 1 V(b) <\sum_{k=1}^{n}\vert \Delta \alpha_k\vert +\frac{\epsilon }{4M}$
and
$\tag 2 \sum_{k=1}^{n}\vert f(t_k')-f(t_k)\vert \vert \Delta \alpha_k\vert <\frac{\epsilon }{4}$.
Now choose the $t'_k\le t_k$ so that
$\sum_{k=1}^{n}\vert M_k(f)-m_k(f)\vert \vert \Delta \alpha_k \vert<\sum_{k=1}^{n}\vert f(t_k')-f(t_k)\vert \vert \Delta \alpha_k \vert+\frac{\epsilon }{4}$.
Then,
$U(P, f, V ) − L(P, f, V )=\sum_{k=1}^{n}(M_k(f)-m_k(f))\Delta V=\sum_{k=1}^{n}(M_k(f)-m_k(f))(\Delta V-\vert \alpha_k\vert)+\sum_{k=1}^{n}(M_k(f)-m_k(f))\vert \alpha_k\vert<$
$2M\sum_{k=1}^{n}(\Delta V_k-\vert\Delta \alpha_k\vert )+\frac{\epsilon }{2}=2M(V(b)-\sum_{k=1}^{n}\vert \Delta \alpha_k \vert)+\frac{\epsilon }{2}<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon,$
which shows that $f$ is R-S integrable with respect to $V$. The second part is now immediate.
edit: to prove 2)., define $S=\left \{ k\in P:\Delta \alpha_k\ge 0 \right \}$ and $T=\left \{ k\in P:\Delta \alpha_k< 0 \right \}$. Then, choose $t_k,t'_k$ so that $f(t_k)-f(t'_k)>M_k(f)-m_k(f)-\epsilon /8V(b)$ whenever $k\in S$, and $f(t'_k)-f(t_k)>M_k(f)-m_k(f)-\epsilon/8V(b)$ whenever $k\in T.\ $
Then,
$\sum_{k=1}^{n}\vert M_k(f)-m_k(f)\vert \vert \Delta \alpha_k \vert<\sum_{k\in S}(f(t_k)-f(t'_k))\vert \Delta \alpha_k\vert+\sum_{k\in T}(f(t'_k)-f(t_k))\vert \Delta \alpha_k\vert+$
$\epsilon /8V(b)\sum_{k=1}^{n}\vert \Delta \alpha_k \vert =\sum_{k=1}^{n}(f(t_k)-f(t'_k))\Delta \alpha_k +$
$\epsilon /8V(b)\sum_{k=1}^{n}\vert \Delta \alpha_k \vert <\epsilon /8+\epsilon /8=\epsilon/4.$
Thus $\tag 3 \sum_{k=1}^{n}\vert f(t_k')-f(t_k) \vert \Delta\alpha_k\vert <\sum_{k=1}^{n}\vert M_k(f)-m_k(f)\vert \vert \Delta \alpha_k \vert<\epsilon /4.$
Best Answer
$F(x)$ must be of bounded variation thats why you couldnt find any counterexample. To see this, observe that if $|f(t)|\leq M$ for every $t\in [\alpha,\beta]$ then, for every $x,y$ \begin{align} |F(x)-F(y)|&=\biggl|\int_{\alpha}^x f(t)\,dt-\int_{\alpha}^y f(t)\,dt\biggr|\\ &=\biggl|\int_{x}^y f(t)\,dt\biggr|\\ &\leq \int_{x}^y |f(t)|\,dt\\ &\leq M|x-y| \end{align} This shows that $F$ is Lipschitz continuous with constant $M$. Now using the definition of bounded variation try to prove that every Lipshcitz continuous function on $[\alpha,\beta]$ is of bounded variation.