Some Product Rule Stuff

Product Rule Extended to More Functions

Let \(f_1,f_2,\ldots,f_n\) be differentiable functions. Then \((f_1f_2\ldots f_n)'=\sum_{i=1}^nf_1^{(\delta_{i1})}f_2^{(\delta_{i2})} \ldots f_n^{(\delta_{in})}\) where \(f_j^{(\delta_{ij})}\) is either \(f_j\) or \(f_j'\) depending on the value of \(\delta_{ij}\), the Kronecker delta, \(\delta_{ij}=\begin{cases}0&i\neq j\\1&i=j\\\end{cases}\).

The base case for \(n=1\), this can be verified. Now assume the theorem is true for \(n\) functions. We will show it is true for \(n+1\) functions. First apply the usual product rule on the \(n+1\) functions:

\[\begin{align}&(f_1f_2\ldots f_nf_{n+1})' =(f_1f_2\ldots f_n)'f_{n+1}+(f_1f_2\ldots f_n)f_{n+1}'\\ &=\left(\sum_{i=1}^nf_1^{(\delta_{i1})}f_2^{(\delta_{i2})}\ldots f_n^{(\delta_{in})}\right)f_{n+1} +(f_1f_2\ldots f_n)f_{n+1}'\\ \end{align}\]

Now \(i\neq n+1\) in the summation. Also, we can combine the other part with the summation since it fits for index \(n+1\). This gives us the desired result:

\[(f_1f_2\ldots f_nf_{n+1})' =\sum_{i=1}^{n+1}f_1^{(\delta_{i1})}f_2^{(\delta_{i2})}\ldots f_n^{(\delta_{in})}f_{n+1}^{(\delta_{i,n+1})}\]

Product Rule Extended to Higher Derivatives

Let \(fg\) be a product of 2 \(n\)-times differentiable functions. Then \((fg)^{(n)}=\sum_{i=0}^n{n\choose i}f^{(i)}g^{(n-i)}\).

The base case \(n=0\) can be shown. Now assume the theorem is true for \(n\) and we will show it for \(n+1\).

\[\begin{align}&(fg)^{(n+1)} =\left(\sum_{i=0}^n{n\choose i}f^{(i)}g^{(n-i)}\right)' =\sum_{i=0}^n{n\choose i}\left(f^{(i+1)}g^{(n-i)}+f^{(i)}g^{(n+1-i)}\right)\\ &=\sum_{i=0}^n{n\choose i}f^{(i+1)}g^{(n-i)} +\sum_{i=0}^n{n\choose i}f^{(i)}g^{(n+1-i)}\\ \end{align}\]

Now shift the indexing of the left summation from \(0,1,\ldots,n\) to \(1,2,\ldots,n\).

\[\begin{align}&=\sum_{i=1}^{n+1}{n\choose i-1}f^{(i)}g^{(n+1-i)} +\sum_{i=0}^n{n\choose i}f^{(i)}g^{(n+1-i)}\\ &=fg^{(n+1)}+\sum_{i=1}^n\left({n\choose i-1}+{n\choose i}\right) f^{(i)}g^{(n+1-i)}+f^{(n+1)}g \\\end{align}\]

Now since \({n\choose i-1}+{n\choose i}={n+1\choose i}\) and we can combine the 2 extra terms to form a summation from \(0\) to \(n+1\) and complete the proof.

\[(fg)^{(n+1)}=\sum_{i=0}^{n+1}{n+1\choose i}f^{(i)}g^{(n+1-i)}\]

For a little extra work, here is proof that \({n\choose i-1}+{n\choose i}={n+1\choose i}\).

\[\begin{align}&\frac{n!}{(i-1)!(n+1-i)!}+\frac{n!}{i!(n-i)!} =\frac{n!\cdot i}{i!(n+1-i)!}+\frac{n!\cdot(n+1-i)}{i!(n+1-i)!}\\ &=\frac{n!(i+n+1-i)}{i!(n+1-i)!}=\frac{n!(n+1)}{i!(n+1-i)!} =\frac{(n+1)!}{i!(n+1-i)!}={n+1\choose i} \end{align}\]