The method
Suppose you wish to calculate the n-th derivative of the product of two functions and . There is a general rule, called Leibniz’s rule, which gives a slightly easier way of computing it, without the process of calculating every step. The rule states that
.
So the n-th derivative is the sum of n+1 terms, with the coefficients given by the n-th line of Pascal’s triangle. This is very much like the binomial theorem, which states that, given two numbers and ,
.
An example
Suppose we wish to calculate . We will still need to calculate the first 5 derivatives of both functions, but that is pretty straight forward. We also calculate the 5th line of Pascal’s triangle.
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So, our derivative is simply multiplying the n-th coefficient with the n-th derivative of and the n-th derivate of , but counting this last one backwards, starting from the bottom of the table.
.
Why does it work?
The idea behind both “rules” is that we apply an operation to a general term, which we shall call the term . The operation is such that:
In particular, we have:
1) In the binomial theorem,
and F is multiplying by :
2) In Leibnitz’s rule,
and F is the derivative operator:
Here, stands for standard addition. We must also state the initial condition,
So, we have the following theorem.
1) If is a vector space and for every non-negative m,n;
2) If there is an operation such that ,
then:
.
This generalizes both rules.
The idea of the proof was found here: http://math.stackexchange.com/questions/135510/how-is-leibnizs-rule-for-the-derivative-of-a-product-related-to-the-binomial-fo .