= Solution
A derivation of $A$ centred at the $k$-point $p:A\to k$ is a $k$-linear map $D:A\to k$ satisfying the twisted <Leibniz rule>
$$
D(ab)=p(a)D(b)+p(b)D(a).
$$
The <Zariski tangent space> is
$$
T_pA=\operatorname{Der}(A,p).
$$
Since $p(f)\ne0$, $p$ extends uniquely to the <localization> $A[f^{-1}]$ by $p(a/f^n)=p(a)/p(f)^n$. Restriction gives a linear map
$$
T_pA[f^{-1}]\longrightarrow T_pA.
$$
Conversely, every $D\in T_pA$ has the unique extension
$$
\widetilde D\!\left(\frac a{f^n}\right)
=\frac{D(a)}{p(f)^n}
-\frac{n\,p(a)D(f)}{p(f)^{n+1}}.
$$
The quotient rule verifies the Leibniz identity, and the assumption that $f$ is not a zero divisor makes the localization representation compatible with equality of fractions. Extension and restriction are inverse, so the natural map is an isomorphism.
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