Leibniz’s theorem

Leibniz’s theorem extends the product rule to determine higher-order derivatives of the product of two or more functions. It can be proven by mathematical induction.

Consider the function $f(x)=u(x)v(x)$ , where $u$ and $v$ are $n$ times differentiable. Using the product rule, the first few derivatives are:

$(uv)'=uv'+u'v$

$(uv)''=uv''+2u'v'+u''v$

$(uv)^{(3)}=uv^{(3)}+3u'v''+3u''v'+u^{(3)}v$

which suggests that the $n$ -th order derivative of $f$ can be expressed as the binomial expansion

$(uv)^{(n)}=\sum_{k=0}^n\begin{pmatrix}n\\k\end{pmatrix}u^{(k)}v^{(n-k)}\;\;\;\;\;\;\;\;319a$

where $n$ and $k$ are non-negative integers, $u^{(j)}$ and $v^{(j)}$ are the $j$ -th order derivatives of $u$ and $v$ , respectively, and $\begin{pmatrix}n\\k\end{pmatrix}\frac{n!}{k!(n-k)!}$ are the binomial coefficients.

Eq319a is known as Leibniz’s theorem, which can be proven by induction. For $n=1$ ,

$(uv)^{(1)}=\begin{pmatrix}1\\0\end{pmatrix}u^{(0)}v^{(1)}+\begin{pmatrix}1\\1\end{pmatrix}u^{(1)}v^{(0)}=uv'+u'v$

If eq319a holds for all $k$ when $n=m$ , then for $n=m+1$ ,

$\begin{align}(uv)^{m+1}&=\frac{d}{dx}\sum_{k=0}^m\begin{pmatrix}m\\k\end{pmatrix}u^{(k)}v^{(m-k)}\\&=\sum_{k=0}^m\begin{pmatrix}m\\k\end{pmatrix}u^{(k+1)}v^{(m-k)}+\sum_{k=0}^m\begin{pmatrix}m\\k\end{pmatrix}u^{(k)}v^{(m-k+1)}\\&=\sum_{k=1}^{m+1}\begin{pmatrix}m\\k-1\end{pmatrix}u^{(k)}v^{(m-k+1)}+\sum_{k=0}^{m}\begin{pmatrix}m\\k\end{pmatrix}u^{(k)}v^{(m-k+1)}\\&=\sum_{k=1}^{m}\begin{pmatrix}m\\k-1\end{pmatrix}u^{(k)}v^{(m-k+1)}+\begin{pmatrix}m\\m\end{pmatrix}u^{(m+1)}v^{(0)}+\begin{pmatrix}m\\0\end{pmatrix}u^{(0)}v^{(m+1)}+\sum_{k=1}^{m}\begin{pmatrix}m\\k\end{pmatrix}u^{(k)}v^{(m-k+1)}\\&=\begin{pmatrix}m+1\\0\end{pmatrix}u^{(0)}v^{(m+1)}+\sum_{k=1}^{m}\biggr\[\begin{pmatrix}m\\k-1\end{pmatrix}+\begin{pmatrix}m\\k\end{pmatrix}\biggr\]u^{(k)}v^{(m-k+1)}+\begin{pmatrix}m+1\\m+1\end{pmatrix}u^{(m+1)}v^{(0)}\end{align}$

Since $\begin{pmatrix}m\\k-1\end{pmatrix}+\begin{pmatrix}m\\k\end{pmatrix}=\begin{pmatrix}m+1\\k\end{pmatrix}$ (see this article for proof),

$\begin{align}(uv)^{m+1}&=\begin{pmatrix}m+1\\0\end{pmatrix}u^{(0)}v^{(m+1)}+\sum_{k=1}^{m}\begin{pmatrix}m+1\\k\end{pmatrix}u^{(k)}v^{(m-k+1)}+\begin{pmatrix}m+1\\m+1\end{pmatrix}u^{(m+1)}v^{(0)}\\&=\sum_{k=0}^{m+1}\begin{pmatrix}m+1\\k\end{pmatrix}u^{(k)}v^{(m-k+1)}\end{align}$

and the theorem holds for all $n$ and $k$ .

Leibniz’s theorem

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