Localization of a differential ring I am reading  Asymptotic Differential Algebra and Model Theory of Transseries and on p. 200 there is a claim/explanation of Localization:

$R$ is a differential ring and $A$ its multiplicative subset, $0 \notin A$.
Then there is a unique derivation on $A^{-1} R$ making $A^{-1} R$ into
a differential ring where the natural map $r \mapsto r / 1: R
\rightarrow A^{-1} R$ is a morphism of differential rings; it is given
by $$ (r / a)^{\prime}=\left(r^{\prime} a-r a^{\prime}\right) / a^2
> \quad \text { for } r \in R, a \in A, $$ and we always consider
$A^{-1} R$ as a differential ring in this way. In particular, if $R$
is a differential integral domain (that is, a differential ring whose
underlying ring is an integral domain), then the derivation $\partial$
of $R$ extends uniquely to a derivation on the fraction field of $R$.

My question is why is the last sentence true? I cannot come up witha proof for it.
EDIT I'm also trying to see why the given derivation actually meets the criteria stated in the first sentence.
The only thing that springs to my mind is universal property of commutative rings (that they use earlier in the book for localization), namely that for every ring morphism $\phi: A \rightarrow B$ into a commutative ring $B$ with $\phi(S) \subseteq B^{\times}$there is a unique ring morphism $\phi^{\prime}: S^{-1} A \rightarrow B$ such that $\phi=\phi^{\prime} \circ \iota$, but I don't see how I could apply it here sicne $R$ a differential algebra.
 A: The quoted claim can be found on the top of page 170, not page 200.
We note the following:

Observation.
Let $R$ be a differential ring, let $r$ be an element of $R$ and let $a$ be an invertible element of $R$.
Then
$$
 r'
 =
 \biggl( \frac{r}{a} ⋅ a \biggr)'
 =
 \biggl( \frac{r}{a} \biggr)' ⋅ a + \frac{r}{a} ⋅ a'
 =
 \biggl( \frac{r}{a} \biggr)' ⋅ a + \frac{r a'}{a} \,,
$$
and therefore
$$
 \biggl( \frac{r}{a} \biggr)'
 =
 \frac{1}{a} \biggl( r' - \frac{r a'}{a} \biggr) \,.
 =
 \frac{r'}{a} - \frac{r a'}{a^2}
 =
 \frac{r' a - r a'}{a^2} \,.
$$

Let now $R$ be a differential ring and let $A$ be a multiplicative subset of $R$.
We need to show that the derivation $∂$ of $R$ extends uniquely to a derivation $D$ on $A^{-1} R$.
In other words, we need to show that there exists a unique derivation $D$ on $A^{-1} R$ with
$$
 D\biggl( \frac{r}{1} \biggr)
 =
 \frac{∂(r)}{1}
$$
for every element $r$ of $R$.
We start by showing the uniqueness of $D$.
We note that for every element $a$ of $A$, the fraction $1 / a$ is invertible in $A^{-1} R$ with inverse given by $a / 1$.
It therefore follows from the above observation that
\begin{align*}
 D\biggl( \frac{r}{a} \biggr)
 &=
 D\biggl( \frac{r / 1}{a / 1} \biggr)
 \\[0.5em]
 &=
 \frac{D(r / 1) ⋅ (a / 1) - (r / 1) ⋅ D(a / 1)}{(a / 1)^2}
 \\[0.5em]
 &=
 \frac{(∂(r) / 1) ⋅ (a / 1) - (r / 1) ⋅ (∂(a) / 1)}{(a^2 / 1)}
 \\[0.5em]
 &=
 \frac{∂(r) a - r ∂(a)}{a^2} \,.
\end{align*}
This shows the uniqueness of $D$.
To show the existence of $D$, we could define it by the above formula, then check that it is well-defined, and finally check that it is again a derivation.
However, we can also use the following useful trick:

Proposition.
Let $$ be a commutative ring and let $A$ be a $$-algebra.
Consider the subalgebra
$$
 \begin{pmatrix}
  A & A \\
  0 & A
 \end{pmatrix}
$$
of $\operatorname{Mat}(2, A)$.
A map
$$
 δ \colon A \to A
$$
is a derivation on $A$ (i.e., $$-linear and satisfying the Leibniz rule $δ(ab) = δ(a) b + a δ(b)$ for all $a, b ∈ A$), if and only if the map
$$
 A
 \to
 \begin{pmatrix}
  A & A \\
  0 & A
 \end{pmatrix} \,,
 \quad
 a
 \mapsto
 \begin{pmatrix}
  a & δ(a) \\
  0 & a
 \end{pmatrix}
$$
is a homomorphism of $$-algebras.

It follows from the above proposition that the map
$$
 R
 \to
 \begin{pmatrix}
  R & R \\
  0 & R
 \end{pmatrix} \,,
 \quad
 r
 \mapsto
 \begin{pmatrix}
  r & ∂(r)  \\
  0 & r
 \end{pmatrix}
$$
is a homomorphism of rings.
We may extend the codomain of this homomorphism, resulting in the homomorphism
$$
 φ
 \colon
 R
 \to
 \begin{pmatrix}
  A^{-1} R  & A^{-1} R \\
  0         & A^{-1} R
 \end{pmatrix} \,,
 \quad
 r
 \mapsto
 \begin{pmatrix}
  r / 1 & ∂(r) / 1  \\
  0     & r / 1
 \end{pmatrix} \,.
$$
The matrix $φ(a)$ is invertible for every element $a$ of $A$, because is an upper triangular matrix with invertible diagonal entries.
More explicitly, we have
$$
 φ(a)^{-1}
 =
 \begin{pmatrix}
  1/a & -∂(a) / a^2 \\
  0   & 1/a
 \end{pmatrix}
$$
for every element $a$ of $A$.
It follows from the universal property of the localization $A^{-1} R$ that $φ$ extends uniquely to a homomorphism of rings
$$
 ψ
 \colon
 A^{-1} R
 \to
 \begin{pmatrix}
  A^{-1} R  & A^{-1} R \\
  0         & A^{-1} R
 \end{pmatrix} \,,
$$
given by
\begin{align*}
 ψ\biggl( \frac{r}{a} \biggr)
 &=
 φ(r) φ(a)^{-1}
 \\
 &=
 \begin{pmatrix}
  r / 1 & ∂(r) / 1 \\
  0     & r / 1
 \end{pmatrix}
 \begin{pmatrix}
  1/a & -∂(a) / a^2 \\
  0   & 1/a
 \end{pmatrix}
 \\[0.5em]
 &=
 \begin{pmatrix}
  r/a & - r ∂(a) / a^2 + ∂(r) / a \\
  0   & r/a
 \end{pmatrix} \,.
\end{align*}
It follows from the above proposition that the map
$$
 D
 \colon
 A^{-1} R
 \to
 A^{-1} R \,,
 \quad
 \frac{r}{a}
 \mapsto
 - \frac{r ∂(a)}{a^2} + \frac{∂(r)}{a}
 =
 \frac{∂(r) a - r ∂(a)}{a^2}
$$
is a derivation on $A^{-1} R$.
We have thus proven the desired existence.
Suppose now that $R$ is an integral domain.
The set $A ≔ R ∖ \{ 0 \}$ is then a multiplicative subset of $R$.
We have seen above that the derivation of $R$ extends uniquely to a derivation on the localization $A^{-1} R$, which is precisely the field of fractions of $R$.
