In the manner of my last post, Irrational Roots, I have been looking over some of my previous work. I have even been able to extend some of it such as that on Finding Multiple Derivatives. What I had been missing in the naiivety was a simple pattern that I dared not approach for fear of complexity. Until now I hadn't the confidence to even explore the possibility of a 'Product on Product Rule,' as I'm choosing to name it now.
My irrational fear was that anything that was more complex the first time would only compound in difficulty over multiple stages. This was not the case and it actually was quite a bit simpler since I had already immersed myself in Binomial Theorem in my work on the General Polynomial Expansion, which meant that I was quite familiar with the pattern that emerged. Firstly I shall give you a look at the pattern that comes about.
f(x) = g(x)h(x)
f'(x) = g'(x)h(x) + g(x)h'(x)
f''(x) = g''(x)h(x) + g'(x)h'(x) + g'(x)h'(x) + g(x)h''(x)
= g''(x)h(x) + 2g'(x)h'(x) + g(x) h''(x)
f'''(x) = g'''(x)h(x) + g''(x)h'(x) + 2g''(x)h'(x) + 2g'(x)h''(x) + g'(x)h''(x) + g(x)h'''(x)
= g'''(x)h(x) + 3g''(x)h'(x) + 3g'(x)h''(x) + g(x)h'''(x)
As my introduction has probably given away, these arrangements are similar to binomial expansions. In fact, they are exactly correlated, as I shall show now. If we were to conduct a similar operation to the binomial expansions for (g'(x) + h'(x))n, where we treat the power on a term as the number of derivatives, we would get the following:
(g'(x) + h'(x))0 = 0C0g(x)h(x)
= g(x)h(x) (notice in the expansion the second term starts with a power of 0 and the first term ends with a power of 0)
(g'(x) + h'(x))1 = 1C0g'(x)h(x) + 1C1g(x)h'(x)
= g'(x)h(x) + g(x)h'(x)
(g'(x) + h'(x))2 = 2C0g''(x)h(x) + 2C1g'(x)h'(x) + 2C2g(x) h''(x)
= g''(x)h(x) + 2g'(x)h'(x) + g(x) h''(x)
(g'(x) + h'(x))3 = 3C0g'''(x)h(x) + 3C1g''(x)h'(x) + 3C2g'(x)h''(x) + 3C3g(x)h'''(x)
= g'''(x)h(x) + 3g''(x)h'(x) + 3g'(x)h''(x) + g(x)h'''(x)
The explanation for this phenomenon is much the same as that for the actual binomial expansions. When the product is applied once, there are all possibilities of combinations between each of the functions, with the sum of the number of derivatives on each in each term being 1. When this is differentiated the possibilities of each term are broken up, thus giving all possibilities with the sum of the number of derivatives being two. Furthermore, the number of times a combination appears is found by ‘choosing’ that number for the derivative from the total number of the derivative for the same reason it is done when doing a binomial expansion.
(g'(x) + h'(x))0 = 0C0g(x)h(x)
= g(x)h(x) (notice in the expansion the second term starts with a power of 0 and the first term ends with a power of 0)
(g'(x) + h'(x))1 = 1C0g'(x)h(x) + 1C1g(x)h'(x)
= g'(x)h(x) + g(x)h'(x)
(g'(x) + h'(x))2 = 2C0g''(x)h(x) + 2C1g'(x)h'(x) + 2C2g(x) h''(x)
= g''(x)h(x) + 2g'(x)h'(x) + g(x) h''(x)
(g'(x) + h'(x))3 = 3C0g'''(x)h(x) + 3C1g''(x)h'(x) + 3C2g'(x)h''(x) + 3C3g(x)h'''(x)
= g'''(x)h(x) + 3g''(x)h'(x) + 3g'(x)h''(x) + g(x)h'''(x)
The explanation for this phenomenon is much the same as that for the actual binomial expansions. When the product is applied once, there are all possibilities of combinations between each of the functions, with the sum of the number of derivatives on each in each term being 1. When this is differentiated the possibilities of each term are broken up, thus giving all possibilities with the sum of the number of derivatives being two. Furthermore, the number of times a combination appears is found by ‘choosing’ that number for the derivative from the total number of the derivative for the same reason it is done when doing a binomial expansion.
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