Question

Intro to generalized arithmetic derivative.

Answer 1

Well, first it starts out like this for any number \[\Large (ab)'=a'b+ab'\] and for primes you have this rule: \[\Large p' =1\] So basically you just apply the first rule as many times as you need until you get down to the prime factorization, really simple. However things like a second derivative are not so easy.

Answer 2

I'll make an example of calculating the first single derivative and first double derivative. Kind of confusing, but once you see it it's not a big deal.

\[\Large 12' = (4*3)' = 4'*3+4*3'\] See, so we can immediately evaluate 3' since it's prime will equal 1. Let's look at 4', we can't say what that is yet, so let's look at it: \[\Large 4' = (2*2)' = 2'*2 + 2*2'\] Now we can evaluate it because it's in terms of primes. Not so bad, so let's substitute these values in: \[\Large 12' = 4*3+4*1 = 16\]

Answer 3

So the double derivative just takes this idea a little further: \[\large 12'' = (2*2*3)'' = (2*2)'' *3 + (2*2)' *3' + 2*2*3''\] See how we have the regular derivative in there as well? The difference here is that the double derivative of two primes multiplied together is 1. So \[\Large (2*2)'' = 1 \\ \Large 3'' = 0 \\ \Large (2*2)' = 2'*2+2*2' = 4\] Notice that last calculation we already did before, it's just the regular single derivative. Plug these all back in and we get: \[\Large 12'' = 3+ 4+0 = 7\]

Now you may think that this definition is totally random, but with one little extra bit of information I can show you how it might be useful to you.