Commuting of derivation on localization and canonical map This problem is from van der Put's "Galois Theory of Linear Differential Equations".
Show that there exists a unique derivation $\partial$ on $RS^{-1}$ (the localization of $R$ with respect to $S$) such that the canonical map for $R \rightarrow RS^{-1}$ commutes with $\partial$ where $R$ is a commutative ring and $S \subset R$ is a multiplicative subset.
Here is my attempt with my rough understanding of the concepts in this problem. Let $\phi : R \rightarrow RS^{-1}$ be the canonical map. We want to show that $\partial (\phi (x)) = \phi (\partial (x))$ for $x \in R$. Observe that by definition, $\phi (\partial (x))$ maps $x \mapsto [\partial (x)],$ the equivalence class of $\partial (x)$ in $RS^{-1}$ (this IS how the canonical map works, right?). On the other hand, $\partial (\phi (x))$ maps $x \mapsto \partial ([x]) = [\partial (x)],$ which is the same equivalence class in $RS^{-1}$ as mapped to by $\phi (\partial (x)).$ Thus, we conclude that $\phi$ and $\partial$ commute. But I'm not sure how to show that there is a unique $\partial$ that satisfies this problem. Could someone please help?
On a tangent: Consider the polynomial ring $R[X_1, X_2, \dots ,X_n ]$ and a multiplicative subset $S \subset R[X_1, X_2, \dots ,X_n ]$. Let $a_1, a_2, \dots , a_n \in R[X_1, X_2, \dots ,X_n ]S^{-1}$ be given. Prove that there exists a unique derivation $\partial $ on $R[X_1, X_2, \dots X_n] S^{-1}$ such that the canonical map $R \rightarrow R[X_1, X_2, \dots ,X_n ] S^{-1}$ commutes with $\partial$ and $\partial (X_i) = a_i$ for all $i$. (Is the assumption $\mathbb{Q} \subset R$ useful at all?)
 A: First, a comment: when we have a derivation of some ring $R$, it is typically a derivation of $R$ as an $A$-algebra for some fixed map $A\to R$, but you have no $A$ in your notation. (We would also require that $\partial(a) = 0$ for all $a\in A$.) This isn't a life-threatening issue, however.
It seems you want to show that there exists a unique derivation $\partial' : S^{-1}R\to S^{-1}R$ (I presume) which commutes with the canonical localization map $\phi$ and a fixed derivation $\partial : R\to R$. I didn't see this original derivation $\partial$ in the statement; I presume it is implicitly fixed. With this set-up, you want to prove that
$$\partial'\circ\phi = \phi\circ\partial.$$
This is almost what you've written (I want to keep $\partial'$ and $\partial$ distinct to avoid confusion). However, you haven't defined the derivation $\partial'$ on $S^{-1}R$! You've shown that for an element of $S^{-1}R$ which is in the image of $\phi$ (call it $\phi(x)$) we must have $\partial'(\phi(x)) = \phi(\partial(x)).$ But what does $\partial'$ do to elements which are not in the image of $\phi$? For example, if $s\in S\setminus R^\times,$ what is $\phi\left(\frac1s\right)$?
To figure this out, let $\partial : R\to R$ be a derivation. Suppose that $\partial' : S^{-1}R\to S^{-1}R$ is a derivation on $S^{-1}R$ such that $\partial'\circ\phi = \phi\circ\partial.$ Let $r/s\in S^{-1}R;$ we want to compute $\partial'(r/s).$ Well, we have
\begin{align*}
\partial'(r/s) &= \partial'\left(r\cdot\frac{1}{s}\right)\\
&= r\partial'\left(\frac{1}{s}\right) + \partial'(r)\frac{1}{s}\\
&= r\partial'\left(\frac{1}{s}\right) + \frac{\partial(r)}{s}.
\end{align*}
So, $\partial'$ is determined by $\partial$ and by what it does on elements of the form $\frac{1}{s}\in S^{-1}R.$ Now, we notice that $\partial'(1) = 0,$ as $\partial'(1) = \partial'(1^2) = 2\partial'(1)$. Thus,
\begin{align*}
0&= \partial'(1)\\
&=\partial'\left(s\cdot\frac{1}{s}\right)\\
&=s\partial'\left(\frac{1}{s}\right) + \frac{\partial(s)}{s}\\
\implies s\partial'\left(\frac{1}{s}\right) &= -\frac{\partial(s)}{s}\\
\implies \partial'\left(\frac{1}{s}\right) &= -\frac{\partial(s)}{s^2},
\end{align*}
which is exactly what we would get if we naively apply the quotient rule from calculus 1.
What we have shown is that if such a derivation $\partial'$ exists, it must be given by the formula $$\partial'\left(\frac{r}{s}\right) = \frac{s\partial(r) - r\partial(s)}{s^2}.$$ This proves uniqueness if such a derivation exists! Now, I leave it to you to check that $\partial'$ as given by this formula is (a) well-defined and (b) a derivation.
Edit: I originally thought the second part of the question wanted $S\subseteq R.$
For your second question, the idea is essentially the same. You need to define $\partial'\left(\frac{f}{g}\right)$ for any $f\in R[x_1,\dots, x_n]$ and $g\in S.$ As above, you can show that you must have
$$
\partial'\left(\frac{f}{g}\right) = \frac{\partial'(f)g - f\partial'(g)}{g^2},
$$
so you simply need to define what $\partial'$ does on elements of $R[x_1,\dots, x_n].$
Now, notice that since any derivation must be linear, it suffices to define $\partial'$ on monomials $rx_1^{m_1}\cdots x_n^{m_n}.$ The Leibniz rule implies that we have
$$
\partial'(rx_1^{m_1}\cdots x_n^{m_n}) = \partial(r)x_1^{m_1}\cdots x_n^{m_n} + r\sum_{i = 1}^n x_1^{m_1}\cdots m_i x_i^{m_i - 1}\cdots x_n^{m_n}\partial(x_i)
$$
(you should verify this if it is not obvious!).
Now, we see that to define $\partial',$ it suffices to define $\partial'(x_i)$ for each $i.$ I leave it to you to show that setting $\partial'(x_i) = a_i$ makes the function $\partial'$ a derivation (no need to assume that $\Bbb{Q}\subseteq R$).
