Presentation of the momentum operator $\hat p$ in $x$-space derived from canonical commutation relation: $[\hat x, \hat p]=\mathrm{i}$.

Here I show how to derive the explicit presentation of the momentum operator $\hat p$ in $x$-space: $\hat p = -\mathrm{i}\frac{d}{dx}$ out of the commutation relation $[\hat x, \hat p]=\mathrm{i}$. I wonder, why all textbooks on quantum mechanics omit this issue. This fact is implicitly kept in mind but never is presented explicitly. The prove of this fact is simple.


Problem:
Let $|x\rangle$ and $|p\rangle$ are the vectors of the abstract physical Hilbert space. Let every vector is uniquely defined by the correspondent eigenvalue of Hermitian operators $\hat x$ and $\hat p$: $\hat x=x|x\rangle$, $\hat p=p|p\rangle$ where $x,p\in\mathbb{R}$. Let $[\hat x, \hat p]=\mathrm{i}$. Prove that $\langle x|\hat p|\psi\rangle = -\mathrm{i}\frac{d}{dx}\langle x|\psi\rangle$. That is prove $\hat p = -\mathrm{i}\frac{d}{dx}$.

Prove:
\begin{equation} [\hat x, \hat p]=\mathrm{i} \end{equation} \begin{equation} \langle x|[\hat x, \hat p]|x_0\rangle=\mathrm{i}\delta(x-x_0) \end{equation} \begin{align} \langle x|\hat x\hat p|x_0\rangle-\langle x|\hat p\hat x|x_0\rangle&=\int\langle x|\hat x|x'\rangle\langle x'|\hat p|x_0\rangle dx' - \int\langle x|\hat p|x'\rangle\langle x'|x_0\rangle x_0 dx'\\ &=x\langle x|\hat p|x_0\rangle - x_0\langle x|\hat p|x_0\rangle = (x-x_0)P(x,x_0) = \mathrm{i}\delta(x-x_0). \end{align} Hence \begin{equation} P(x,x_0) = \mathrm{i}\frac{\delta(x-x_0)}{(x-x_0)} \end{equation} where $P(x,x_0) = \langle x|\hat p|x_0\rangle$.

Now:
\begin{align} \langle x|\hat p|\psi\rangle &= \mathrm{i}\int\langle x|\hat p|x_0\rangle\langle x_0|\psi\rangle = \mathrm{i}\int\frac{\delta(x-x_0)}{(x-x_0)}\psi(x_0) dx_0 = -\mathrm{i}\int\delta'(x-x_0)\psi(x_0) dx_0\\ &= \mathrm{i}\int\delta'(x_0-x)\psi(x_0)dx_0 = -\mathrm{i}\frac{d}{dx}\psi(x)
\end{align} where properties of the Dirac delta function were used.

Q.E.D.