### Derivatives of a data set

The basic idea of a derivative amounts to finding the change in one variable with respect to another.

Here is an example. Let's take some data for a freely falling object. It starts falling at t=0, and we record its position every second after. We can put this data into two columns in Excel, then plot distance as a function of time. We see the expected parabolic shape.

Let's say we now want to figure out the derivative of position with respect to time: $\frac{dy}{dt}$. If we knew the functional relationship between y and t, this would be easy. $$y = -\frac{1}{2}gt^2 \Rightarrow \frac{dy}{dt}=-gt$$. However, we are working with data points, and not a function. Thus, we have to do a numerical derivative. We can't actually set our $\Delta t \rightarrow 0$, so we'll have to approximate it.

We'll have to manually figure out the change in height (our y variable), with respect to time. Something like the following should suffice. $$y' = \frac{\Delta y}{\Delta t} = \frac{y_1-y_0}{t_1-t_0}$$

Below is a picture showing this calculation in the Excel sheet. The cell D3 is calculated using the formula =(B3-B2)/(A3-A2). All the subsequent rows in that column are just the same formula applied to next rows down. The cells in column C are just the midpoints between the times used in the calculation.

We are essentially finding the slope between each point in the position graph, then plotting those slopes as a function of time. Since the slope of the position w.r.t. time is velocity, we now have a velocity graph. For example, between 10 and 11 seconds, the position changed by -104.2 m. This tells us the velocity at that point (t = 10.5) will be approximately -104.2/1 = 104.2 meters, as shown in the zoom in below. Looking at the velocity graph, we can see that this indeed is the case. The velocity at 10.5 is about 100 m/s.

Lastly, we would like to find the slope of the velocity graph in order to determine the free-fall acceleration of this object. For this, we can use the linear fit tool, as shown below. Indeed, we recover an expected 9.8 m/s2.

This is just the most basic way to perform a numerical derivative. There are more thorough methods which you may explore later on.