In my recent post, Finding Multiple Derivatives: The Product On Product Rule, I have continued my work on further differentiation to extend it to the product rule. For the purpose of finding a general rule for extending this to the chain rule (or function of a function rule), I have found a extension for the product rule, a 'Multiple Products Rule,' if you like. What I mean by this is that, instead of finding the derivative of two functions multiplied together, we find the derivative for the product of multiple functions. The well known garden variety product rule is given by:
If f(x) = g(x)h(x),
f'(x) = g'(x)h(x) + g(x)h'(x)
One might expect that the general rule would be a similar sum of products where each function has its derivative multiplied by the other functions. So, it is as I shall show now.
Assuming true for f(x) = g(x)h(x)i(x)...,
f'(x) = g'(x)h(x)i(x)... + g(x)h'(x)i(x)... + g(x)h(x)i'(x)... + ...
To prove for e(x)f(x),
To prove for e(x)f(x),
Show that for d(x) = e(x)f(x),
d'(x) = e'(x)g(x)h(x)i(x)... + e(x)g'(x)h(x)i(x)... + e(x)g(x)h'(x)i(x)... + e(x)g(x)h(x)i'(x)... + ...
d'(x) = e'(x)f(x) + e(x)f'(x)
= e'(x)(g(x)h(x)i(x)...) + e(x)(g'(x)h(x)i(x)... + g(x)h'(x)i(x)... + g(x)h(x)i'(x)... + ...)
= e'(x)g(x)h(x)i(x)... + e(x)g'(x)h(x)i(x)... + e(x)g(x)h'(x)i(x)... + e(x)g(x)h(x)i'(x)... + ...
∴ If true for f(x), it is true for e(x)f(x), c(x)e(x)f(x), and so on for the product of an infinite number of functions
Since for 1xF(x), F'(x) = 1xF'(x) + 0xF(x) = F'(x), it is true for all such products.
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