A good explanation for Leibniz's rule for higher derivatives is similar to the Binomial theorem(in general, multinomia theorem)

Leibniz's rule for higher derivatives is similar to the Binomial theorem.

$\begin{array}{c}{\left(\frac{d}{dx}\right)}^n(uv)=\sum _{i=0}^n(\genfrac{}{}{0ex}{}{n}{k})\left(u^{(n)}{v}^{(n-k)}\right)\end{array}$$\left(\frac{d}{dx}\right)^n(uv)=\sum_{i=0}^n \binom{n}{k}\left(u^{(n)}v^{(n-k)}\right)\\$

But why?

This article will give a natural approach.

Consider $\begin{array}{c}\frac{duv}{dx}=\frac{du}{dx}v+u\frac{dv}{dx}\end{array}$$\frac{duv}{dx}=\frac{du}{dx} v+u\frac{dv}{dx}\\$

The differential operator is like ${\partial }_1+{\partial }_2$$\partial_1+\partial_2$ acts on $uv$$uv$

${\partial }_i$$\part_i$ mean derivative for the $i^{th}$$i^{th}$ function.

Thus $\begin{array}{c}\frac{duv}{dx}=({\partial }_1+{\partial }_2)(uv)\end{array}$$\frac{duv}{dx}=(\partial_1+\partial_2)(uv)\\$, and both $\begin{array}{c}\frac{d}{dx},({\partial }_1+{\partial }_2)\end{array}$$\frac{d}{dx},(\partial_1+\partial_2)\\$ is linear

So $\begin{array}{c}{\left(\frac{d}{dx}\right)}^n(uv)\end{array}$$\left(\frac{d}{dx}\right)^n(uv)\\$

$\begin{array}{c}=({\partial }_1+{\partial }_2{)}^n(uv)=\sum _{i=0}^n(\genfrac{}{}{0ex}{}{n}{k})({\partial }_1{)}^n({\partial }_2{)}^{n-k}(uv)=\sum _{i=0}^n(\genfrac{}{}{0ex}{}{n}{k})\left(u^{(n)}{v}^{(n-k)}\right)\end{array}$$=(\partial_1+\partial_2)^n(uv)=\sum_{i=0}^n \binom{n}{k}(\part_1)^n(\partial_2)^{n-k}(uv)=\sum_{i=0}^n \binom{n}{k}\left(u^{(n)}v^{(n-k)}\right)\\$

And more generally, we can consider $\begin{array}{c}\frac{duvw}{dx}=({\partial }_1+{\partial }_2+{\partial }_3)(uvw)\end{array}$$\frac{duvw}{dx}=(\partial_1+\partial_2+\partial_3)(uvw)\\$

And $\begin{array}{c}\frac{d\prod _{i=1}^n{u}_i}{dx}=(\sum _{i=1}^n{\partial }_i)(\prod _{i=1}^n{u}_i)\end{array}$$\frac{d\prod_{i=1}^n u_i}{dx}=(\sum_{i=1}^n\partial_i)(\prod_{i=1}^n u_i)\\$

To see this, we just need to consider symmetry

According to associative law, we have $\begin{array}{c}\frac{duvw}{dx}=\frac{d(uv)w}{dx}=\frac{duv}{dx}w+uv\frac{dw}{dx}\end{array}$$\frac{duvw}{dx}=\frac{d(uv)w}{dx}=\frac{duv}{dx}w+uv\frac{dw}{dx}\\$

We do not need to expand it; just consider it commutative, $uvw=vwu=wuv$$uvw=vwu=wuv$

$u,v,w$$u,v,w$ is symmetry, so it must equal to $\begin{array}{c}\frac{du}{dx}vw+u\frac{dv}{dx}w+uv\frac{dw}{dx}\end{array}$$\frac{du}{dx}vw+u\frac{dv}{dx}w+uv\frac{dw}{dx}\\$

So $\begin{array}{c}\frac{duvw}{dx}=({\partial }_1+{\partial }_2+{\partial }_3)(uvw)\end{array}$$\frac{duvw}{dx}=(\partial_1+\partial_2+\partial_3)(uvw)\\$

And then, we see $\begin{array}{c}\frac{d\prod _{i=1}^n{u}_i}{dx}=(\sum _{i=1}^n{\partial }_i)(\prod _{i=1}^n{u}_i)\end{array}$$\frac{d\prod_{i=1}^n u_i}{dx}=(\sum_{i=1}^n\partial_i)(\prod_{i=1}^n u_i)\\$ for the same reason.

Thus we can apply the multinomial theorem to it.

$\begin{array}{c}(a_1+a_2+\dots +a_m{)}^n=\sum _{k_1+k_2+\dots +k_m=n}(\genfrac{}{}{0ex}{}{n}{k_1,k_2,\dots ,k_m})\cdot (a_1^{k_1}\cdot a_2^{k_2}\cdot \dots \cdot a_m^{k_m})\end{array}$$(a_1 + a_2 + \ldots + a_m)^n = \sum_{k_1+k_2+\ldots+k_m=n} \binom{n}{k_1,k_2,\ldots,k_m} \cdot (a_1^{k_1} \cdot a_2^{k_2} \cdot \ldots \cdot a_m^{k_m})\\$