Lucas Willems


A 25 year-old student passionate about maths and programming


Nth derivative of product


The nth derivate of product of 2 functions is given by Leibniz' formula :

$$(fg)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} f^{(n-k)} g^{(k)}$$

where \(f\) et \(g\) are 2 functions \(n\) times derivable, \(f^{(l)}\) means \(l\)-th derivate of \(f\) and \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).

Just like Newton's binomial formula, this formula is easily conjecturable, but much more difficult to prove. The purpose of this article is to show you how to prove it.



If you know Newton's binomial formula, you will notice that these 2 formulas (Newton's and Leibniz') are very similar, because they "work" in the same way : induction is the same.

So, we have to use induction with this statement :

$$\forall n \in \mathbb{N} \qquad H_n : (fg)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} f^{(n-k)} g^{(k)}$$that leads us to the following reasoning :

Bases :
For \(n = 0\), \((fg)^{(0)} = fg = \binom{0}{0} f^{(0)} g^{(0)}\). So, \(H_0\) holds.

Induction steps :
For \(n+1\) :

$$(fg)^{(n+1)} = \left((fg)^{(n)}\right)'$$

As we assume \(H_n\) holds, we have :

$$\begin{align} (fg)^{(n+1)} &= \left(\sum_{k=0}^{n} \binom{n}{k} f^{(n-k)} g^{(k)}\right)' = \sum_{k=0}^{n} \binom{n}{k} \left(f^{(n-k)} g^{(k)}\right)' \\ &= \sum_{k=0}^{n} \binom{n}{k} \left(f^{(n-k+1)} g^{(k)}+f^{(n-k)} g^{(k+1)}\right) \\ &= \sum_{k=0}^{n} \binom{n}{k} f^{(n-k+1)} g^{(k)} + \sum_{k=0}^{n} \binom{n}{k} f^{(n-k)} g^{(k+1)} \\ &= \sum_{k=0}^{n} \binom{n}{k} f^{(n-k+1)} g^{(k)} + \sum_{k=1}^{n+1} \binom{n}{k-1} f^{(n-(k-1))} g^{((k-1)+1)} \\ &= \binom{0}{0}f^{(n+1)}f^{(0)} + \sum_{k=1}^{n} \binom{n}{k} f^{(n+1-k)} g^{(k)} + \binom{n}{n}f^{(0)}g^{(n+1)} + \sum_{k=1}^{n} \binom{n}{k-1} f^{(n+1-k)} g^{(k)} \\ &= \binom{0}{0}f^{(n+1)}g^{(0)} + \binom{n+1}{n+1}f^{(0)}g^{(n+1)} + \sum_{k=1}^{n} \left(\binom{n}{k} + \binom{n}{k-1}\right)f^{(n+1-k)} g^{(k)} \end{align}$$

As Pascal's triangle allows us to see easily that \(\binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}\), we get :

$$\begin{align} (fg)^{(n+1)} &= \binom{0}{0}f^{(n+1)}g^{(0)} + \binom{n+1}{n+1}f^{(0)}g^{(n+1)} + \sum_{k=1}^{n}\binom{n+1}{k}f^{(n+1-k)} g^{(k)} \\ &= \sum_{k=0}^{n+1}\binom{n+1}{k}f^{(n+1-k)}g^{(k)} \end{align}$$

Furthermore, \(H_n \Rightarrow H_{n+1}\).

Conclusion :
\(H_n\) holds \(\forall n \in \mathbb{N}\).

We have just proved the formula of nth derivative of product of 2 functions !


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