With all honesty I find myself in difficulty to understand how the rules of switching in quantum mechanics work. I need these to understand and to prove the Heisemberg uncertainty principle. For example, with the rules of commutation, I would like to prove (myself) that

$$\Delta p_y \cdot\Delta y \geq \frac{\hslash}2\tag{1}$$

My starting point is: $[A,B]=AB-BA \tag{a}$

Now, why

$$\left[\frac{\partial }{\partial x},x\right]\stackrel{?}{=}1 \tag{2}$$ I have thought, from the rule (a),

$$\left[\frac{\partial }{\partial x},x\right]=\frac{\partial }{\partial x}x-x\frac{\partial }{\partial x}=1-\ldots$$

This identity $$\left[x,\frac{\partial }{\partial x}\right]=-1$$ is easy because $[A,B]=-[B,A]$.

After, why, $$\left[i\hslash\frac {\partial}{\partial x},x\right]=i\hslash \tag{3}$$

$$[p_x,x]=i\hslash \tag{4}$$

I have also these information that I have not understood:

\begin{equation} \frac{\partial(xf(x))}{\partial x}-x\frac{\partial f(x)}{\partial x}=f(x)+x \frac{\partial f}{\partial x}-x\frac{\partial f}{\partial x}=f(x) \end{equation}

$$\left[x,\frac{\partial }{\partial x}\right]\cdot f(x)=\frac{\partial(xf(x))} {\partial x}-x\frac{\partial f(x)}{\partial x}=f(x)$$

In other words, you can help me understand how I will get the proof of the Heisemberg uncertainty principle. They are very appreciate simple books where I can find very simple operations with these commutators. I have never studied quantum mechanics.

I thank in advance anyone who could help me.