The Radon-Nikodym “derivative” is an a.e. define concept. Suppose $(X,S)$ is a measure space and $\mu,\nu$ are finite measures on $(X,S)$ with $\mu\ll\nu$, then the theorem is:
Theorem. There exists $f\in L^{1}(X,\nu)$ a non-negative real-valued function, with $\mu(A)=\int_{x\in A} f(x)~\nu(dx)$ for all $A\in S$.
There are all sorts of generalisations (to $\sigma$-finite, signed, and complex-valued measures, etc.). This is the theorem/lemma in its simplest form. It is easy to show, that if $f,f'$ satisfy the theorem, then $f'=f$ $\nu$-a.e. (and thus also $\mu$-a.e.). Thus the function is “unique” warranting the definite article.
Since it effectively follows, that $\int g~d\mu=\int gf~d\nu$ for $g$ in $L^{1}(X,\mu)$, on may apply the nomenclature "$d\mu=f d\nu$", so that the notation $\frac{d\mu}{d\nu}$ for $f$ is an intuitive name. Since $f$ has the said property, it follows, that it is an “effective” derivative.
Here an outline of the proof/construction:
Lemma. Let $\sigma$ be a signed mass on $X$ (this means, countably additive and finite). Then there is a measurable set $A\subseteq X$ satisfying $(\forall{B\in S})~B\subseteq A\Rightarrow\sigma(B)\geq 0$
and $(\forall{B\in S})~B\subseteq X\setminus A\Rightarrow\sigma(B)\leq 0$. Thus $\sigma=\sigma^{+}-\sigma^{-}$, whereby $\sigma^{+}=\sigma\mid_{A}$ and $\sigma^{-}=\sigma\mid_{X\setminus A}$ are non-negative finite measures with disjoint support. Moreover, any such decomposition of $\sigma$ is of this form and $A$ is unique $\sigma$-a.e.
Proof of Theorem. (Sketch). Let $A_{r}\in S$ be as in the Lemma for the signed measure $\sigma_{r}:=r\nu-\mu$ for each $r\in\mathbb{Q}$. Consider now two rationals $r<r'$. Then $\sigma_{r'}=\sigma_{r}+(r'-r)\nu$. Let $B=A_{r}\setminus A_{r'}$. It holds per definition of the decomposition(s), that
$$0\geq\sigma_{r'}(B)=\sigma_{r}(B)+(r'-r)\nu(B)\geq 0+(r'-r)\nu(B),$$
thus $\nu(B)=0$. Thus $A_{r}\subseteq^{\ast}_{\nu}A_{r'}$. Replacing $A_{r}$ with the thus $\nu$-a.e. equivalent set $\bigcup_{q\in\mathbb{Q},q\leq r}A_{q}$, it can be assumed that $(A_{q})_{q\in(\mathbb{Q},<)}$ ist a monotone increasing family of measurable sets.
Consider finally the map $f:x\in X\mapsto \inf\{r\in\mathbb{Q}:x\in A_{r}\}\in[-\infty,\infty]$. This can readily be shown to be well-defined, measureable and $\nu$-a.e. $>\infty$. Let $B\in S$. For all reals, $r<r'$, it holds that $f^{-1}[r,r')=A_{r'}\setminus A_{r}$ $\nu$-a.e. and thus also $\mu$-a.e.; it therefore holds that
\begin{align}
\int_{B\cap f^{-1}[r,r')}f~d\nu-\mu(B\cap f^{-1}[r,r'))
&\geq r\nu(B\cap A_{r'}\setminus A_{r})-\mu(B\cap A_{r'}\setminus A_{r})\\
&= (r-r')\nu(B\cap A_{r'}\setminus A_{r})
+\sigma_{r'}(B\cap A_{r'}\setminus A_{r})\\
&\geq -(r'-r)\nu(B\cap A_{r'}\setminus A_{r}+0\\
&= -(r'-r)\nu(B\cap f^{-1}[r,r')).\\
\end{align}
and
\begin{align}
\int_{B\cap f^{-1}[r,r')}f~d\nu-\mu(B\cap f^{-1}[r,r'))
&\leq r'\nu(B\cap A_{r'}\setminus A_{r})-\mu(B\cap A_{r'}\setminus A_{r})\\
&= (r'-r)\nu(B\cap A_{r'}\setminus A_{r})
+\sigma_{r}(B\cap A_{r'}\setminus A_{r})\\
&\leq (r'-r)\nu(B\cap A_{r'}\setminus A_{r}+0\\
&= (r'-r)\nu(B\cap f^{-1}[r,r')).\\
\end{align}
Thus
$$\left|\int_{B\cap f^{-1}[r,r')}f~d\nu-\mu(B\cap f^{-1}[r,r'))\right| \leq (r'-r)~\nu(B\cap f^{-1}[r,r')).$$
It follows that, decomposing $B$ into countably many preimages under $f$ of sufficiently small intervals of the real line, that
$$\left|\int_{B}f~d\nu-\mu(B)\right|\leq\varepsilon\nu(B)$$
for all $\varepsilon>0$. Thus $\int_{B}f~d\nu=\mu(B)$ for all $B\in S$ (in particular $f\in L^{1}(\nu)$).
Q.e.d.
Interpreting the derivative (aside from its application).
The construction in the proof allows one to describe the derivative. It holds for $\varepsilon>0$ and reals $a<b$ with $|b-a|<\varepsilon$ and $\nu$-a.e. $x\in X$ that $a\leq\frac{d\mu}{d\nu}(x)<b$ if and only if $x\in A_{b}\setminus A_{a}=:B$, which means the set of such $x$ constitutes a set $B$ with $b\nu(B)-\mu(B)=\sigma_{b}(B)\geq 0$ and $a\nu(B)-\mu(B)=\sigma_{a}(B)\leq 0$.
It immediately follows, that either $\frac{d\mu}{d\nu}$ is $\nu$-a.e. not in $[a,b)$, or else $a\leq\frac{\mu(B)}{\mu(B)}\leq b$. Thus for $|b-a|<\varepsilon$ small
$$\left|\frac{d\mu}{d\nu}(x)-\frac{\mu(B)}{\nu(B)}\right|<\varepsilon$$
for $\nu$-a.e. $x\in B:=\{y\in X:\frac{d\mu}{d\nu}(y)\in[a,b)\}$.
Or in other words: let $f=\frac{d\mu}{d\nu}$. Then for $\nu$-a.e. $x\in X$ it holds that $\frac{d\mu}{d\nu}(x)=\lim_{\varepsilon\to 0^{+}}\frac{\mu(f^{-1}U_{\varepsilon,x})}{\nu(f^{-1}U_{\varepsilon,x})}$, where $U_{\varepsilon,x}\subseteq\mathbf{R}$ the $\varepsilon$-interval around $f(x)\in\mathbf{R}$.
