Let $X$ be an affine algebraic variety and $A=\mathbb{C}[X]$ its coordinate ring. The vector space $$ \mathrm{Der}_{\mathbb{C}}(A) = \{ D: A \rightarrow A \mid D \text{ is linear and } D(ab)=D(a)b+aD(b) \ \forall a,b \in A \} $$ is an $A$-module. Let $S$ be a multiplicative closed set on $A$, then we may consider the $S^{-1}A$-module $S^{-1} \mathrm{Der}_{\mathbb{C}} (A) $.
Is it true that $S^{-1}\mathrm{Der}_{\mathbb{C}}(A) \cong \mathrm{Der}_{\mathbb{C}}(S^{-1} A)$?
If yes, how do I prove it? Furthermore, is it true for other associative algebras?
The short answer to your narrow question is yes. The slightly longer answer is that, for any commutative ring $k$ and finitely presented commutative $k$-algebra $A$, $S^{-1} \textrm{Der}_k (A) \cong \textrm{Der}_k (S^{-1} A)$. The question doesn't make sense if $A$ is non-commutative: in that case $\textrm{Der}_k (A)$ has no obvious $A$-module structure so what would $S^{-1} \textrm{Der}_k (A)$ even be?
We could prove the claim directly, but it is somewhat tedious because the universal property of $S^{-1} A$ is with respect to $A$-linear maps or ring homomorphisms but derivations are neither. It is more convenient to use a different perspective here.
Definition. Given a ring $A$ and an $A$-bimodule $M$, the Beck module $A \oplus M \epsilon$ is the ring defined as follows:
There is an obvious ring embedding $A \to A \oplus M \epsilon$, sending each $a$ to $a + 0 \epsilon$, and there is also a surjective ring homomorphism $A \oplus M \epsilon \to A$ sending each $a + m \epsilon$ to $a$. However, there can be many other ring embeddings $A \to A \oplus M \epsilon$, not necessarily sending $a$ to $a + 0 \epsilon$. Indeed:
Proposition. Let $k$ be a commutative ring, let $A$ be a $k$-algebra, let $M$ be an $A$-bimodule, and let $D : A \to M$ be a $k$-linear map. The following are equivalent:
The map $\phi_D : A \to A \oplus M \epsilon$ defined by $\phi_D (a) = a + D (a) \epsilon$ is a $k$-algebra homomorphism.
$D : A \to M$ is a derivation, i.e. $D (a_0 a_1) = D (a_0) a_1 + a_0 D (a_1)$ for all $a_0$ and $a_1$ in $A$.
Proof. Suppose $\phi_D$ as defined is a $k$-algebra homomorphism. Then, $$\phi_D (a_0 a_1) = \phi_D (a_0) \phi_D (a_1) = a_0 a_1 + D (a_0) a_1 \epsilon + a_0 D (a_1) \epsilon$$ but $\phi_D (a_0 a_1) = a_0 a_1 + D (a_0 a_1) \epsilon$ as well, hence $D$ is indeed a derivation.
Conversely, if $D$ is a derivation then reversing the above argument shows $\phi_D$ preserves multiplication. It is clear that $\phi_D$ is $k$-linear and preserves $1$, so we indeed have a $k$-algebra homomorphism. ◼
In other words:
Corollary. Let $k$ be a commutative ring, let $A$ be a $k$-algebra, and let $M$ be an $A$-bimodule. There is a natural bijection between the set of $k$-linear derivations $A \to M$ and the set of $k$-algebra homomorphisms $\phi : A \to A \oplus M \epsilon$ such that $\pi \circ \phi = \textrm{id}_A$, where $\pi : A \oplus M \epsilon \to A$ is the $k$-algebra homomorphism defined by $\pi (a + m \epsilon) = a$. ◼
From here on, $A$ will be commutative and $A$-bimodules will be symmetric (i.e. $a m = m a$). Note that $A \oplus M \epsilon$ is then also commutative, and the $k$-module $\textrm{Der}_k (A, M)$ of $k$-linear derivations $A \to M$ has an induced $A$-module structure.
Proposition. Let $S$ be a multiplicatively closed subset of $A$. The restriction operation $\textrm{Der}_k (S^{-1} A, S^{-1} M) \to \textrm{Der}_k (A, S^{-1} M)$ is bijective, i.e. for every $k$-linear derivation $D : A \to S^{-1} M$ there is a unique $k$-linear derivation $\tilde{D} : S^{-1} A \to S^{-1} M$ such that $\tilde{D} \left( \frac{a}{1} \right) = D (a)$.
Proof. Suppose given a $k$-linear derivation $D : A \to S^{-1} M$. Let $\phi_D : A \to S^{-1} A \oplus S^{-1} M \epsilon$ be the map defined by $\phi_D (a) = \frac{a}{1} + D (a) \epsilon$. It is straightforward to verify that $\phi_D$ is a $k$-algebra homomorphism. Furthermore, if $s \in S$, then $\phi_D (s)$ is a unit in $S^{-1} A \oplus S^{-1} M \epsilon$: indeed, $$\left( \frac{s}{1} + D (s) \epsilon \right) \left( \frac{1}{s} - \frac{1}{s^2} D (s) \epsilon \right) = \frac{1}{1} + 0 \epsilon$$ so there is a unique $k$-algebra homomorphism $\tilde{\phi} : S^{-1} A \to S^{-1} A \oplus S^{-1} M \epsilon$ such that $\tilde{\phi} \left( \frac{a}{1} \right) = \phi_D (a)$ for all $a \in A$. We have $\pi \circ \tilde{\phi} = \textrm{id}_{S^{-1} A}$, hence $\tilde{\phi}$ corresponds to a $k$-linear derivation $\tilde{D} : S^{-1} A \to S^{-1} M$ such that $\tilde{\phi} \left( \frac{a}{s} \right) = \frac{a}{s} + \tilde{D} \left( \frac{a}{s} \right)$. It is clear that $\tilde{D} \left( \frac{a}{1} \right) = D (a)$, and it is straightforward to check that the construction $D \mapsto \tilde{D}$ defined here is the required two-sided inverse for the restriction operation. ◼
Proposition. There is a well-defined map $S^{-1} \textrm{Der}_k (A, M) \to \textrm{Der}_k (A, S^{-1} M)$ sending each formal fraction $\frac{D}{s}$ to the derivation $a \mapsto \frac{D (a)}{s}$, and it is bijective if $A$ is finitely presented.
Proof. There is an obvious $A$-linear map $\textrm{Der}_k (A, M) \to \textrm{Der}_k (A, S^{-1} M)$ induced by the natural map $M \to S^{-1} M$. Since $S$ acts invertibly on $\textrm{Der}_k (A, S^{-1} M)$, this immediately extends to an $S^{-1} A$-linear map $S^{-1} \textrm{Der}_k (A, M) \to \textrm{Der}_k (A, S^{-1} M)$, and it is easy to check that it is the map in the proposition.
It remains to be shown that the map is bijective if $A$ is finitely presented. There is an abstract nonsense proof, which basically amounts to observing that $S^{-1} M$ is a colimit of a filtered diagram that $\textrm{Der}_k (A, -)$ maps to the diagram whose colimit defines $S^{-1} \textrm{Der}_k (A, M)$, and when $A$ is finitely presented then $\textrm{Der}_k (A, -)$ preserves filtered colimits, so we are done. Spelling all this out concretely gives a direct proof: essentially, given a $k$-linear derivation $D : A \to S^{-1} M$ and a finite presentation of $A$, we can find some $s \in S$ that is "big" enough so that there is a well-defined map $A \to M$ defined on the generators by $a \mapsto s D (a)$. ($s$ needs to be "big" enough not just to clear the denominator of $D (a)$ for each generator $a$, it also needs to annihilate all the relations between the chosen generators!) ◼
Corollary. If $A$ is a finitely presented commutative $k$-algebra and $S$ is a multiplicatively closed subset of $A$, then $S^{-1} \textrm{Der}_k (A) \cong \textrm{Der}_k (S^{-1} A)$. ◼
Since there seem to be doubts about whether a finiteness hypothesis is required, here is a counterexample. Let $A = k [s, t_0, t_1, t_2, \ldots]$ and let $S = \{ 1, s, s^2, \ldots \}$. There is a unique $k$-algebra homomorphism $A \to S^{-1} A \oplus S^{-1} A \epsilon$ sending $s$ to $s$ and each $t_n$ to $t_n + \frac{t_n}{s^n} \epsilon$. This extends to a $k$-algebra homomorphism $S^{-1} A \to S^{-1} A \oplus S^{-1} A \epsilon$, so we get a $k$-linear derivation $D : S^{-1} A \to S^{-1} A$. It has the property that (even after reducing fractions to lowest terms) there is no constant $N$ such that $s^N D (a) \in A$ for all $a \in A$. Hence, there is no $k$-linear derivation $\tilde{D} : A \to A$ such that $D = \frac{1}{s^N} \tilde{D}$ for some $N$, demonstrating the non-surjectivity of $S^{-1} \textrm{Der}_k (A) \to \textrm{Der}_k (S^{-1} A)$.