Mathematics in selection of RC time constant for an envelope detector for AM

Let the capacitor be charged to voltage $V(t_0)$ at $t=t_0$ and then start discharging. At time $t \, (>t_0)$, the delayed capacitor voltage is $$v(t)=V(t_0)\operatorname{exp}(- \frac{t-t_0}{RC}). \tag 1$$

The rate of change of $v(t)$ AT $t=t_0$ is $$\frac{dv}{d(t-t_0)}|_{t=t_0}=-\frac{V(t_0)}{RC}. \tag 2$$

My questions:

1. When they are saying rate of change of $v(t)$ , why have they differentiated $v(t)$ with respect to $(t-t_0)$, why haven't they differentiated with respect to $t$ and then subtituted $t$ with $t_0$ , which gives the same answer as shown: \begin{align} \frac{dv}{dt} &= V(t_0)e^{\frac{t_0}{R_C}}\frac{d}{dt}(e^{-\frac{t}{RC}}) \\ &=V(t_0)e^{\frac{t_0}{RC}}e^{-\frac{t}{RC}}(-\frac{1}{RC}) \\ &= >\frac{dv}{dt}|_{t=t_0}=-\frac{V(t_0)}{RC} \end{align}

2. How to differentiate $v(t)$ w.r.t $(t-t_0)$. I am particularly confused about this because of the component $V(t_0)$ of eq $(1)$. Should I consider it a constant or should I consider it a variable as I am differentiating $v(t)$ with respect to $(t-t_0)$. In other words how to perform this differentiation?

• Isn't $d(t-t_0) = dt$ since $t_0$ is a constant? – Atul Ingle Jul 20 '17 at 20:32
• Yes thats true!!! – Soumee Jul 20 '17 at 20:35
• @AtulIngle Then why are they complicating such a simple thing? Should I differentiate it just with.r.t $t$ instead of $t-t_0$. I wanted to know if anything in mathematics is there wherein they differentiate a function wrt $t-t_0$, but as you pointed out both are just the same in case $t_0$ is a constant, I would like to know what will be the differentiating procedure if $t_0$ is a variable. – Soumee Jul 20 '17 at 20:41
• I have no idea what the physical meaning of "variable $t_0$" would be in this example. That aside, if $t_0$ were a function of $t$ you can work out its differential in the usual way. Eg. if $t_0=t^2$, then $dt_0 = 2t dt$. – Atul Ingle Jul 20 '17 at 21:37
• No answer as to why except to point out that the decay is proportional to the time difference; but for calculations, you just apply the chain rule from calculus. – rrogers Jul 25 '17 at 19:36