Is there any significance to the derivative of the average? I have a set of time vs. concentration data that I'm trying to turn into an average derivative value for use in an equation (I'm worried a simple average of the first and last points ignores too much data). The most obvious way to do it would be to simply do $\frac{C - C_0}{t - t_0}$ for each point and take the average, but I'm wondering, is there any significance to the value $\frac{C_{avg}}{t_{avg}}$? Since I have uniform time intervals I think this is equivalent to 
$$\frac{1}{N} \sum_{i}{\frac{C_i}{\Delta}}$$
where $\Delta$ is the time interval length and $N$ is the number of samples. If possible, avoiding numerical differentiation would be great since I'm getting some weird values from that approach. Any advice would be greatly appreciated. Thanks!
 A: The derivative of the average value of the function on a range depends on both the range of the function and the function itself.
$$\text{avg of $f(x)$ on $x\in[a,b]$ of }=\frac1{b-a}\int_a^bf(t)dt$$
Let's consider how to calculate an average given the average of the elements before it. First, the discrete version where you only add one element to a set:
$$\overline f(x+1)=\frac{f(x+1)+n\overline f(x)}{n+1}$$
where $\overline f(x)$ is the average of the first $x$ elements and $n$ is the number of elements in the set before adding the new element. We can rewrite this in terms of a continuous interval:
$$\overline f(x+h)=\frac{hf(x+h)+(b-a)\overline f(x)}{b-a+h}$$
Finally, we can find $\frac d{dx}\overline f(x)$.
$$\begin{align}\frac d{dx}\overline f(x)&=\lim_{h\to0}\frac1h(\overline f(x+h)-\overline f(x))\\
&=\lim_{h\to0}\frac1h\left(\frac{hf(x+h)+(b-a)\overline f(x)}{b-a+h}-\overline f(x)\right)\\
&=\lim_{h\to0}\frac1h\left(\frac{hf(x+h)+(b-a)\overline f(x)-(b-a+h)\overline f(x)}{b-a+h}\right)\\
&=\lim_{h\to0}\frac1h\left(\frac{hf(x+h)-h\overline f(x)}{b-a+h}\right)\\
&=\lim_{h\to0}\frac{f(x+h)-\overline f(x)}{b-a+h}\\
&=\frac{f(x)-\overline f(x)}{b-a}
\end{align}
$$
That's the explicit formula for what you're looking for. I do not know the name of this idea yet.
