Let me try to answer the question how I understand it (basically, just to expand a bit on the comments by Harald and Willie).

Let $Df$ be a distributional derivative of the differentiable function $f:\mathbb R\to\mathbb R$. This implies, in particular, that $Df$ is a linear continuous functional on the space of test functions $\mathcal{D}(\mathbb R)$. By definition, $Df$ can be identified with a a function $g:\mathbb R\to\mathbb R$ iff $$\langle Df,\phi\rangle=(g,\phi)\equiv\int_{\mathbb R} \phi(x)g(x)dx\qquad\qquad\qquad\qquad \quad(1)$$ for all $\phi\in\mathcal{D}(\mathbb R)$. If such $g$ exists, it is unique (up to its modifications on a measure zero set). Now, we want to know when $g$ exists and is equal to the classical derivative $f'$ of $f$.

• First, in order for the r.h.s. of (1) to be finite for all $\phi\in\mathcal{D}(\mathbb R)$, the function $g=f'$ must be locally integrable on $\mathbb R$ in the sense of Lebesgue, i.e. $$\int\limits_{a}^{b}|g(x)|dx=\int\limits_{a}^{b}|f'(x)|dx < \infty, \quad \forall a,b\in\mathbb R.\qquad(2)$$ In other words, the function $f$ should have bounded variation on all finite intervals $[a,b]\in\mathbb R$. Note, that $f'$ is always measurable so (2) simply means that the derivative is not too "large".

• Condition (2) is necessary but not sufficient as the example in Rudin's book shows. Indeed, the measure $d\mu(x)=|f'(x)|dx$ may fail to be absolutely continuous with respect to the Lebesgue measure $dx$. This is the case when $d\mu$ has jumps, e.g. check out the singular measure determined by Cantor's function as Willie suggested).

• Finally, if $f$ is absolutely continuous on $\mathbb R$ then $Df$ can be identified with $f'$ as Rudin explains (roughly speaking, one can justify the integration by parts in (1) in this case). This is a necessary and sufficient condition.

Let me try to answer the question how I understand it (basically, just to expand a bit on the comments by Harald and Willie).

Let $Df$ be a distributional derivative of a differentiable function $f:\mathbb R\to\mathbb R$. This implies, in particular, that $Df$ is a linear continuous functional on the space of test functions $\mathcal{D}(\mathbb R)$. By definition, $Df$ can be identified with a function $g:\mathbb R\to\mathbb R$ iff $$\langle Df,\phi\rangle=(g,\phi)\equiv\int_{\mathbb R} g(x)\phi(x)dx\qquad\qquad\qquad\qquad \quad(1)$$ for all $\phi\in\mathcal{D}(\mathbb R)$. If such $g$ exists, it is unique (up to modifications on a measure-zero set).

So the question is: given a differentiable function $f$, when does (1) hold with $g=f'$?

• First, in order for the integral in (1) to be finite for all $\phi\in\mathcal{D}(\mathbb R)$, the function $g=f'$ must be locally integrable on $\mathbb R$ in the sense of Lebesgue, i.e. $$\int\limits_{a}^{b}|g(x)|dx=\int\limits_{a}^{b}|f'(x)|dx < \infty, \quad \forall a,b\in\mathbb R.\qquad(2)$$ In other words, the function $f$ should have bounded variation on all finite intervals $[a,b]\subset\mathbb R$. Note, that $f'$ is always measurable so (2) simply means that the derivative is not wildly unbounded.

• Condition (2) is necessary but not sufficient. If $f$ is of bounded variation then by Lebesgue's decomposition theorem it can be written as $$f=f_{ac}+f_{sing}$$ where $f_{ac}$ is absolutely continuous and $f_{sing}$ is a step function. The example in Rudin's book shows that if $f_{sing}$ does not vanish then we cannot integrate by parts to get the identity $$\int_{\mathbb R} f'(x)\phi(x)dx=-\int_{\mathbb R} f(x)\phi'(x)dx\qquad\qquad\qquad(3)$$ and (1) also fails.

• Finally, if $f$ is absolutely continuous on $\mathbb R$ (i.e. $f=f_{ac}$) then $Df$ can be identified with $f'$ (one can justify the integration by parts in (3) and get (1) with $g=f'$ in this case). This is a necessary and sufficient condition.