I'm reading Kähler Differentials written by Ernst Kunz. There is an exercise in its first chapter: Let $R$ be a $K$-algebra and $\delta:R\to R$ a $K$-derivation. Let $N\subset R[x]$ be multiplicatively closed and $\alpha\in N^{-1}R[x].$ There is precisely one derivation $\delta_\alpha: N^{-1}R[x]\to N^{-1}R[x]$ with $\delta_{\alpha}(x)=\alpha$ such that the below diagram
is commutative. I have extended $\delta$ to $\bar{\delta}$ and $d$, which the first one is determined uniquely, but my problem is with showing that $d$ is unique.
Let $\bar{\delta}:R[x] \to N^{-1}R[x]$ be an extension of $\delta$ that maps $x$ to $\alpha$, and $\bar{\delta}(r)=\dfrac{\delta r}{1}$ (for $r\in R$). Image of an arbitrary member of $R[x]$ can be uniquely determined by the definition of $\bar{\delta}$, using the Leibniz rule. And the below digram is commutative:
Let $d: N^{-1}R[x]\to N^{-1}R[x]$ be an extension of $\bar{\delta}$: $d(\dfrac{f_1}{f_2}):=(f_2\bar{\delta}(f_1)-f_1\bar{\delta}(f_2))(f_2^2)^{-1}$. Where the inverse is taken in $N^{-1}R[x]$; $d$ satisfies the Leibniz rule and the below diagram is commutative:
Edit: Thanks for the answer
Let $r=\dfrac{f_1}{f_2}\in N^{-1}R[x]$, we have $r\dfrac{f_2}{1}=\dfrac{f_1}{1}$, $\Rightarrow d(r\dfrac{f_2}{1})=d(\dfrac{f_1}{1})$. Since $d(\dfrac{f_i}{1})=\bar{\delta}(f_i)$, using Leibniz rule we have $\dfrac{f_2}{1}d(r)+r\bar{\delta}(f_2)= \bar{\delta}(f_1)$. $\Rightarrow d(r)=(\bar{\delta}(f_i)-r\bar{\delta}(f_2))(f_2)^{-1}$; Hence for all $r\in N^{-1}R[x]$, $d(r)$ is determined uniquely.
Consider a fraction $f=g/h\in N^{-1}R[x]$. Then $hf=g$, so $d(g)=d(hf)=fd(h)+hd(f)$ for any derivation $d:N^{-1}R[x]\to N^{-1}R[x]$. But now since $h$ acts as a unit on $N^{-1}R[x]$, we can uniquely solve for $d(f)$: we must have $$d(f)=h^{-1}(d(g)-fd(h))=h^{-2}(hd(g)-gd(h)).$$ So, $d$ is uniquely determined by what it does on elements that come from $R[x]$.