Given a function x(t) and a constant T, is the following relation true?: $$\frac{dx(t-T)}{d(t-T)}=\frac{dx(t-T)}{dt}$$
If it is true, what is the proof?
Also, if this is true, does this mean that this statement is wrong?: $$\frac{dx(t-T)}{d(t-T)}=\frac{dx(z)}{dz}=\frac{dx(t)}{dt}$$ where $z=t-T$.
Here is a snippet from Example 1.11 of Linear Systems and Signals by B. P. Lathi and Roger Green using this relation:
Applying input $x(t)$ to the system produces output $y(t) = \frac{d}{dt}x(t)$; delaying this output by T produces $y(t − T) = \frac{d}{d(t−T)}x(t − T) = \frac{d}{dt}x(t − T)$. This is just the output of the system to a delayed input $x(t−T)$. Since the T-delayed output of the system to input $x(t)$ equals the output of the system to the T-delayed input $x(t−T)$, the system is time invariant.
This part refers to proving the system $y(t)=\frac{d}{dt}x(t)$ to be time invariant.
The proof is the chain rule, using $$\frac{d(t-T)}{dt} = 1\qquad\text{or its consequence}\qquad \frac{dt}{d(t-T)} = 1$$
Yes, $$ \frac{dx(z)}{dz}=\frac{dx(t)}{dt} $$ is wrong, where $z=t-T$. Unless $x(t)$ is periodic with period $T$, then it is right.