# LUCAS WILLEMS

A 25 year-old student passionate about maths and programming

# Nth derivative of product

## Article

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.

Summary

## Proof

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 !