Calculating derivatives of products using Pascal’s Triangle

The method

Suppose you wish to calculate the n-th derivative of the product of two functions f and g. 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

\displaystyle (fg)^{(n)}=\sum_{k=0}^n\dbinom{n}{k}f^{(k)}g^{(n-k)}.

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 a and b,

\displaystyle \left(a+b\right)^n=\sum_{k=0}^n\dbinom{n}{k}a^{k}b^{n-k}.

An example

Suppose we wish to calculate \left(e^x \sin x\right)^{(5)}. 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.

n (e^x)^{(n)} (\sin x)^{(n)}
1 e^x \cos x
2 e^x -\sin x
3 e^x -\cos x
4 e^x \sin x
5 e^x \cos x
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

So, our derivative is simply multiplying the n-th coefficient with the n-th derivative of e^x and the n-th derivate of \sin x, but counting this last one backwards, starting from the bottom of the table.

(e^x\sin x)^{(5)}= \\ \\ =1e^x(\cos x)+5e^x(\sin x)+10e^x(-\cos x)+10e^x(-\sin x)+5e^x(\cos x)+1e^x(\sin x)= \\ =e^x(\cos x+5\sin x-10\cos x-10\sin x+5\cos x+\sin x)= \\ \\ =-4e^x(\sin x +\cos x).

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 T(n,m). The operation F is such that:


In particular, we have:
1) In the binomial theorem,
T(n,m)=a^nb^m and F is multiplying by (a+b):

2) In Leibnitz’s rule,
T(n,m)=f^{(n)}g^{(m)} and F is the derivative operator:
F( T(n,m) )=\frac{d}{dx}(f^{(n)}g^{(m)})=f^{(n+1)}g^{(m)}+f^{(n)}g^{(m+1)}=\\ T(n+1,m)+T(n,m+1)

Here, + stands for standard addition. We must also state the initial condition,
(a+b)^0=1\\ (fg)^{0}=fg

So, we have the following theorem.
1) If V is a vector space and T(n,m)\in V for every non-negative m,n;
2) If there is an operation F such that F(~T(n,m)~)=T(n+1,m)+T(n,m+1),
\displaystyle F^n(T(0,0))=\sum_{k=0}^{n}\dbinom{n}{k}T(k,n-k).

This generalizes both rules.

The idea of the proof was found here: .


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