Let $X:[0,T]\to V$ be a continuous path taking values in some Banach space $V$.
The $p$-variation is defined as, with the supremum being take over all partitions of $[0,T]$,
$$\|X\|_p:= \sup_D \left(\sum_{t_j \in D} |X_{t_j}-X_{t_{j-1}}|^p\right)^{\frac{1}{p}}$$
If $X$ is $C^1$, it is well-known that the $1$-variation (also known as the length) is equal to the $L^1$ norm of the first derivative.
Is it the case that for such $X$, the $L^2$-norm of the first derivative (sometimes known as the energy) equals the $2$-variation? More generally, is the $p$-th norm of the first derivative equal to $p$-th variation?
So, here is a counterexample that shows that the $p$-th norm of the first derivative is not equal to the $p$-variation for $p>1$, take the function $X \colon \left[0,2\right] \to \mathbb{R}$ defined as $X\left(t\right) = t$, so
$\lVert X\rVert_{p} \geq \left(\left|2-0\right|^{p}\right)^{\frac{1}{p}} = 2 > 2^{\frac{1}{p}} = \left(\int_{0}^{2} \left|1\right|^{p}dt\right)^{\frac{1}{p}} = \left(\int_{0}^{2} \left|X'\left(t\right)\right|^{p}dt\right)^{\frac{1}{p}}$.
I don't know if you can find an expression for the $p$-variation that is related to the derivative. If I had to guess I would say yes, but I didn't find any reference for this.