Does differentiable function of bounded variation have bounded derivative?

Solution 1:

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 $[a_n,b_n]$.

Solution 2:

$f(x)=\sqrt {x}$ is function of bounded variation but its derivative is unbounded.