Finding Multiple Derivatives: The General Product Rule

          As some readers may have picked up already, in my last post, Further Differentiation: Multiple Products Rule, I missed a very exciting next step that will combine it with my work on both Finding Multiple Derivatives, and the General Polynomial Expansion.  This fantastic insight is a General Product Rule for Multiple Derivatives.

        By following the reasoning in my posts on the Multiple Products Rule, and the post Finding Multiple Derivatives: The Product On Product Rule, it arises that for further differentiation of multiple products, the result will be similar to the polynomial expansion of the sum of the multiplicands to the power of the order of derivative.  If in a term a function has a certain power, that power should become the order of derivative on that function within that term rather than a power in accordance with my previous work.

      Examples

(1)          f(x) = g(x)h(x)i(x)

(2)         f'(x) = g'(x)h(x)i(x) + g(x)h'(x)i(x) + g(x)h(x)i'(x)

(3)         f''(x) = g''(x)h(x)i(x) + g'(x)h'(x)i(x) +g'(x)h(x)i'(x) + g'(x)h'(x)i(x) + g(x)h''(x)i(x) + g(x)h'(x)i'(x) + g'(x)h(x)i'(x) + g(x)h'(x)i'(x) + g(x)h(x)i''(x)

                      = g''(x)h(x)i(x) + 2g'(x)h'(x)i(x) + 2g'(x)h(x)i'(x) + g(x)h''(x)i(x) + 2g(x)h'(x)i'(x) +  + g(x)h(x)i''(x)

                    (by the Multiple Product Rule)

          Alternatively,

(1)          (g + h + i)⁰ = g⁰h⁰i⁰
(2)          (g + h + i)¹ = g¹h⁰i⁰ + g⁰h¹i⁰ + g⁰h⁰i¹
(3)          (g + h + i)² = g²h⁰i⁰ + 2g¹h¹i⁰ + 2g¹h⁰i¹ + g⁰h²i⁰ + 2g⁰h¹i¹ + g⁰h⁰i²
                     (by Polynomial Theorem)

∴    (1)    f(x) = g(x)h(x)i(x)
      (2)    f'(x) = g'(x)h(x)i(x) + g(x)h'(x)i(x) + g(x)h(x)i'(x)
      (3)   f''(x) = g''(x)h(x)i(x) + 2g'(x)h'(x)i(x) + 2g'(x)h(x)i'(x) + g(x)h''(x)i(x) + 2g(x)h'(x)i'(x) +  + g(x)h(x)i''(x)
                     (General Product Rule)