Let $f:[0,1]\rightarrow\mathbb{R}$ be a continuous function, differentiable on $(0,1)$ and such that $\,f'$ is continuous on $(0,1)$. Prove that $f$ is of bounded variation and $$TV(f,[0,1]) = \int_0^1 \left|f'\right|d\mu$$
I prove that $\,f'$ is measurable and the integral is well defined, but I can't prove that $f$ is of bounded variation, mainly because I can't bound the supremum of $|f'|$.
The given premises don't guarantee that $f$ has bounded variation. A counterexample is, as so often, built on the topologists sine curve. Let
$$f(x) = \begin{cases}\quad 0 &, x = 0\\ x\sin \frac1x &, x \neq 0. \end{cases}$$
$f$ is continuous (on all of $\mathbb{R}$), differentiable on $\mathbb{R}\setminus\{0\}$, and its derivative
$$f'(x) = \sin \frac1x - \frac1x\cos \frac1x$$
is continuous there. But the integral of $\lvert f'\rvert$ is not finite. The $\sin\frac1x$ part is bounded, hence harmless, so let's look at the other part. For $k \in \mathbb{Z}$, we have $\cos t \geqslant \frac12$ for $a_k := 2k\pi - \frac\pi3 \leqslant t \leqslant 2k\pi + \frac\pi3 =: b_k$, so
$$\begin{align} \int_0^1 \left\lvert\frac1x\cos\frac1x\right\rvert\,dx &\geqslant \sum_{k=1}^\infty \int_{1/b_k}^{1/a_k} \left\lvert \frac1x \cos \frac1x\right\rvert\, dx\\ &\geqslant \sum_{k=1}^\infty \frac12 \int_{1/b_k}^{1/a_k} \frac{dx}{x}\\ &= \frac12 \sum_{k=1}^\infty \log \frac{b_k}{a_k}\\ &= \frac12 \sum_{k=1}^\infty \log \frac{1 + \frac{1}{6k}}{1-\frac{1}{6k}}\\ &\geqslant \sum_{k=1}^\infty \frac{1}{6k}\\ &= +\infty. \end{align}$$