Laurent Schwartz had the idea. Distributions are linear functionals defined on test functions. They generalize functions in this sense. If $u$ is an actual function, then the corresponding distribution is the linear functional
$$
\phi \mapsto \int u \;\phi\;dx
$$
A linear functional not of this form may still be considered a generalized function.
Now if $u$ is a function and the derivative $v = D^\alpha u$ actually exists, then integration by parts shows
$$
\int_{\Omega}v\;\phi\,dx = (-1)^{|\alpha|}\int_{\Omega} u \,D^{\alpha}\phi\;dx
$$
[no boundary terms because $\phi$ vanishes outside a compact subset of $\Omega$]
Next, if $u$ s a function but $D^\alpha u$ does not exist in the classical sense, it is still true that the functional
$$
\phi \mapsto (-1)^{|\alpha|}\int_{\Omega} u\; \,D^{\alpha}\phi\;dx
$$
makes sense and defines a generalized function (a.k.a.Schwartz distribution). So things work out if we go ahead and call this functional $D^\alpha u$. [It is not a function, but a distribution.]
As I said in the comment: I figured it out and here is the proof in case somebody else has the same problem in the future.
Proof:
For $1 \leq k \leq n$ we define $v_k$ inductively:
Set $v_n:=u^n$. A theorem which was proven earlier in my lecture states that there exists a function $\Phi_1 \in \mathcal{C}^\infty_c(a,b)$ with $\int_a^b \Phi_1(x) \mathrm{d}x=1$.
For all $ 1 \leq k <n$ and $x \in (a,b)$ we define
$$v_k(x):=\int_a^x v_{k+1}(y) \mathrm{d}y+\underbrace{(-1)^k \int_a^b u(z)\Phi_1^k(z) \mathrm{d}z-\int_a^b (\int_a^z v_{k+1}(y) \mathrm{d}y)\Phi_1(z)\mathrm{d}z.}_{=:c} $$
We claim that for all $1 \leq k <n$: $v_k \in L^1(a,b)$ is well defined, a weak derivative of order k of u, and differentiable almost everywhere, i.e., absolutely continuous, with $v_k^\prime=v_{k+1}$.
We prove this by induction over $n-k$:
Assume that $v_{k+1} \in L^1(a,b)$ is well defined and a weak derivative of order k+1 of u.
Firstly, we will show that $x \mapsto \int_a^x v_{k+1}(y) \mathrm{d}y \in L^1(a,b)$:
$$\int_a^b |\int_a^x v_{k+1}(y) \mathrm{d}y|\mathrm{d}x \leq \int_a^b \int_a^x |v_{k+1}(y)| \mathrm{d}y \mathrm{d}x\leq \int_a^b \underbrace{\int_a^b |v_{k+1}(y)| \mathrm{d}y}_{<\infty,\ v_{k+1} \in L^1(a,b)} \mathrm{d}x<\infty.$$
This implies that $v_k$ is well defined and that $v_k \in L^1(a,b)$. By the fundamental theorem of calculus we know that $v_k$ is differentiable almost everywhere with weak derivative $v_k^\prime=v_{k+1}$.
Let $\Phi \in \mathcal{C}^\infty_c(a,b)$. There is a theorem that states that there exist $d \in \mathbb{R}$ and $\Phi_0 \in \mathcal{C}^\infty_c(a,b)$ s.t. $\Phi=\Phi_0^\prime + d\Phi_1$. This implies
$$(-1)^k \int_a^b v_k(x)\Phi(x)\mathrm{d}x=(-1)^k \int_a^b v_k(x) (\Phi_0^\prime(x)+d\Phi_1(x)) \mathrm{d}x=(-1)^k \int_a^b v_k(x)\Phi_0^\prime(x) \mathrm{d}x+(-1)^k d \int_a^b v_k(x) \Phi_1(x)\mathrm{d}x \underset{v_k^\prime=v_{k+1}}{=} (-1)^{k+1} \int_a^b v_{k+1}(x)\Phi_0(x) \mathrm{d}x+(-1)^k d \int_a^b (\int_a^x v_{k+1}(y) \mathrm{d}y +c)\Phi_1(x) \mathrm{d}x \underset{\text{def. of c and }\int \Phi_1=1}{=} (-1)^{k+1} \int_a^b v_{k+1}(x)\Phi_0(x) \mathrm{d}x + (-1)^k d (-1)^k \int_a^b u(x)\Phi_1^k(x)\mathrm{d}x \underset{v_{k+1} \text{ weak deriv. of order k+1}}{=}\int_a^b u(x)\Phi_0^{k+1}(x) \mathrm{d}x+\int_a^b u(x)d\Phi_1^k(x)\mathrm{d}x=\int_a^b u(x) \underbrace{(\Phi_0^\prime+d\Phi_1)^k}_{=\Phi^k}(x) \mathrm{d}x.$$
This means that $v_k$ is a weak derivative of order k of u and thereby finishes the proof.
Best Answer
It's not true. Well, one can always define a distributional derivative: for any $f \in L^1_{loc} (\Omega)$, define $Df$ as a functional $Df : C_c^\infty (\Omega) \to \mathbb R$ given by
$$Df[\phi] := - \int_\Omega f \phi' .$$
So what you are asking is whether or not a distributional derivative (a functional) is represented by an function $v\in L^1(\Omega)$
$$Df [\phi]= \int_\Omega v \phi, \ \ \ \ \forall \phi$$
(Note if this is true, we call $v$ the weak derivative and $f$ is said to be in the Sobolev space $W^{1, 1}(\Omega)$).
There are continuous function $f$ on $(-1, 1)$ so that $Df$ is not a function. A lot of example is suggested in
https://mathoverflow.net/questions/38751/a-h%C3%B6lder-continuous-function-which-does-not-belong-to-any-sobolev-space
even for Holder continuous function $f$.