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When Gottfried Wilhelm von Leibniz worked out his version of the calculus he derived the rules governing the process of differentiation (and introduced the notation that we use today). Among those rules we find a theorem concerning multiple differentiations of a product of two functions. We have before us the task of deducing the n-th derivative, with respect to the variable x, of the product of the functions f(x) and g(x).
Start with the basic definition of a derivative,
Now we need to apply that definition to obtaining the first derivative of the product f(x)g(x). Substituting that product into Equation 1 gives us
If we calculate out the product [f(x+dx)-f(x)][g(x+dx)-g(x)], we get
Substitute the right side of that equation for its left side in Equation 2. Dividing that by dx and taking the limit yields a finite result for the first three terms in that expression and an ignorably minuscule result for the fourth term. We get that result because the first three terms are singly minuscule, as is the divisor dx, and the fourth term is doubly minuscule. So now we know that
That procedure could get extremely tedious for multiple differentiations, so we want to find an alternate method for working out a formula for calculating a multiple derivative of a product of two functions.
Equation 4 tells us that applying a differentiation operator to a product results in a sum in which each term consists of the derivative of one factor multiplied by the other factor. Each of those derivatives becomes a factor in a new product, so applying a second differentiation to the product above yields the product of the derivatives of both factors plus both ways of obtaining one of the factors multiplied by the second derivative of the other. That pattern persists for all differentiations of the original product, so the n-th derivative of the product consists of n+1 terms consisting of every possible application of the n-fold differentiation operator to the product,
in which P(n,k) represents the coefficient of the k-th term in the sum.
To determine the algebraic form of P(n,k) we start with the product of one factor with the n-th derivative of the other factor and derive a description of the next term in the sum as we transfer the differentiation operation from one factor to the other. In how many different ways can we do that? To answer that question we simply count the number of different ways in which we can put n pebbles labeled d/dx into two bowls labeled f(x) and g(x). We have only one way to put the pebbles into one bowl; we have n ways to put one pebble into one bowl and the rest into the other bowl; we have n(n-1) ways to put two pebbles into one bowl and the rest into the other bowl; and so on. We thus discern that P(n,k)=n!/k!(n-k)!, the (k+1)-th binomial coefficient of order n. Substituting that result into Equation 5 then gives us,
Thus we get Leibniz痴 rule of multiple differentiation of a product of two functions, an expression that looks like the formula that expresses the binomial theorem.
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