Graphing the derivative with just data points?

[Sig00]

Hello everyone, my professor didn't have time to teach this so I'm kind of on my own, and so far Google has failed me. But I'm wondering how I can graph the derivative of just data points. I was given one hundred different points of data and told to graph the derivative of them. Maybe I'm missing something obvious? I've graphed the original equation, but by the sounds of the follow up problems after graphing it he's going to need us to be pretty precise so I guess no estimating. I don't know, any help would be super appreciated!

t (days) P (insects)
0 10
1 11
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4 15
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11 29
12 32
13 35
14 39
15 43
16 47
17 52
18 57
19 63
20 69
21 75
22 83
23 91
24 99
25 109
26 119
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29 154
30 167
31 182
32 197
33 213
34 231
35 249
36 268
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100 995

 

[HallsofIvy]

With just "data points" you can't get the derivative which, using a limit, requires knowing all values near the point of differentiation. You can, of course, approximate the derivative. There are several different ways to do that.
The simplest way is to just use the "difference quotient", the slope of the straight line between consecutive points. For example, you have, at one point, "(15, 43), (16, 47), (17, 52)". The line from (15, 43) to (16, 47) has slope (47- 43)/(16- 15)= 4/1= 4 and the line from (16, 47) has slope (52- 47)/(17- 16)= 5/1= 5. Or you could average those two, saying that the derivative at x= 16 is (4+ 5)/2= 4.5.

Given three points there exist a unique quadratic giving those points. We can write the quadratic as \(\displaystyle y= a(x- 16)^2+ b(x- 16)+ c\) and use the three data points to determine the three values, a, b, and c. Writing "x- 16", because the middle point has x= 16, simplifies the calculation: at x= 15, \(\displaystyle 43= a(-1)^2+ b(-1)+ c= a- b+ c\), at x= 16, \(\displaystyle 47= a(0)^2+ b(0)+ c= c\), and at x= 17, \(\displaystyle 52= a(1)^2+ b(1)+ c= a+ b+ c\). So we have a- b= 43- 47= -4 and a+ b= 52- 47= 5. Then 2a= 1 so a= 1/2 and 2b= 5+ 4= 9 so b= 9/2. The quadratic through those three points is \(\displaystyle y= (1/2)(x- 16)^2+ (9/2)(x- 16)+ 47\). The derivative of that is \(\displaystyle y'= (x- 16)+ 9/2\) so that \(\displaystyle y'(16)= 1+ 9/2= 11/2= 5.5.

Or you could take four consecutive points, find the cubic through those four points and differentiate. One would hope that the more work done the more "accurate" the derivative but without knowing the "true" function it is impossible to be certain. Also, while having more data points normally makes numerical computations more accurate, because the difference quotient goes to "0/0", with more data points, the derivative may become "unstable".\)