Why General Leibniz rule and Newton's Binomial are so similar?

Here's one combinatorial way to look at both formulas.

For the first one, let $M$ be the operator which eats a polynomial $f(x,y)$ and returns the polynomial $(x+y)f(x,y)$. Note $M$ is linear, since multiplication is distributive. We want to start with $1$, apply $M$ $n$ times, and see what we get. The point is that $M$ sends any monomial $x^r y^s$ to a sum of two related ones: $$M : x^r y^s \to x^{r+1}y^s + x^r y^{s+1}.$$

Therefore, $(x+y)^n$ enumerates paths from the top of the following diagram to the bottom row; the coefficient of $x^k y^{n-k}$ is the number of paths from $1$ to $x^k y^{n-k}$, but that's $\binom nk$ since any path is a sequence of $n$ left/right choices, and you have to go left $k$ times and right $n-k$ times.

$$\matrix{ &&&&&&1\\ &&&&&\swarrow&&\searrow\\ &&&&x&&&&y\\ &&&\swarrow&&\searrow&&\swarrow&&\searrow\\ &&x^2&&&&xy&&&&y^2\\ &&&&\dots&&\dots&&\dots\\ &\swarrow&&\searrow&&\swarrow&&\searrow&&\swarrow&&\searrow\\ x^n&&&&x^{n-1}y&&\dots&&xy^{n-1}&&&&y^n }$$

For the second formula, $D$ is the derivative operator; we would like to apply it $n$ times to the product $f g$. Notice that $D$ acts nicely on $f^{(r)}g^{(s)}$: $$D : f^{(r)}g^{(s)}\to f^{(r+1)}g^{(s)}+f^{(r)}g^{(s+1)}.$$

So $(fg)^{(n)}$ enumerates paths from the top of the following diagram to the bottom row. $$\matrix{ &&&&&&f^{(0)}g^{(0)}\\ &&&&&\swarrow&&\searrow\\ &&&&f^{(1)}g^{(0)}&&&&f^{(0)}g^{(1)}\\ &&&\swarrow&&\searrow&&\swarrow&&\searrow\\ &&f^{(2)}g^{(0)}&&&&f^{(1)}g^{(1)}&&&&f^{(0)}g^{(2)}\\ &&&&\dots&&\dots&&\dots\\ &\swarrow&&\searrow&&\swarrow&&\searrow&&\swarrow&&\searrow\\ f^{(n)}g^{(0)}&&&&f^{(n-1)}g^{(1)}&&\dots&&f^{(1)}g^{(n-1)}&&&&f^{(0)}g^{(n)} }$$

Same graph, same numbers of paths, same coefficients.

Suppose, we have two functions $ f(x) $ and $ g(x)$ , then the Taylor expansion of both around a point $ x= \alpha $ is of the form:

$$ f(x) = \sum_{k=0}^{\infty} a_k (x-\alpha)^k$$

$$ g(x) = \sum_{k=0}^{\infty} b_k (x-\alpha)^k$$

and, if we multiply the two series, we get another series of the form,

$$ f(x) \cdot g(x) = \sum_{k=0}^{\infty} c_k (x-\alpha)^k$$

where, $$ c_k = \sum_{i=0}^{k} a_k b_{k-i}$$

by the Cauchy product formula, and now, we bring in Taylor's theorem,

$$ a_i = \frac{f^{i} (a)}{i!} $$

and,

$$ b_{k-i} =\frac{ g^{k-i} (a) }{(k-i)!}$$

And,

$$ c_k = \sum_{i=0}^{k} \frac{f^{i} (a)}{i!} \frac{ g^{k-i} (a) }{(k-i)!}$$

Now, suppose, we take definition of $c_k$ from the definition of Taylor expansion, then,

$$ c_k = \frac{ (fg)^k }{k!}\bigg|_{a}$$

Equating the two definitions for $ c_k$

$$ \frac{ (fg)^k }{k!}\bigg|_{a} = \sum_{i=0}^{k} \frac{f^{i} (a)}{i!} \frac{ g^{k-i} (a) }{(k-i)!}$$

re-arranging

$$ (fg)^k\bigg|_{a} = \sum_{i=0}^{k} \binom{k}{i} f^{i} (a) g^{k-i} (a)$$

At this point, we can just replace '$a$' with '$x$' because we are not really evaluating anything using the definitions of functions yet, also, we replace '$k$' with '$n$' for aesthetic

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

Note: functions are assumed to be nice Note: exponents on functions mean derivatives not powers.

Taylor series expansion of $f(x+h)$ $$ f(x+h)=f(x)+h\frac{d}{dx} \left(f(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x) \right)+.... $$

Taylor series expansion of $g(x+h)$ $$ g(x+h)=g(x)+h\frac{d}{dx} \left(g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( g(x) \right)+.... $$

Taylor series expansion of $f(x+h)g(x+h)$ $$ f(x+h)g(x+h)=f(x)g(x)+h\frac{d}{dx} \left(f(x)g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x)g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x)g(x) \right)+.... $$

---

$$ f(x+h)g(x+h)=f(x)g(x)+h\frac{d}{dx} \left(f(x)g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x)g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x)g(x) \right)+....=(f(x)+h\frac{d}{dx} \left(f(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x) \right)+....)(g(x)+h\frac{d}{dx} \left(g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( g(x) \right)+....) $$

If you order $h^n$, it will give you binom expansion

$$\frac{d^{n}}{dx^{n}} f(x)g(x)=\sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x)$$

Because it has exactly the same coefficient of

$$ =(1+hx +\frac{h^2 x^2 }{2!}+\frac{h^3x^3 }{3!}+....)(1+hy +\frac{h^2 y^2 }{2!}+\frac{h^3y^3 }{3!}+....)=e^{hx}e^{hy}=e^{h(x+y)} $$

$$ e^{h(x+y)}=1+h(x+y) +\frac{h^2 (x+y)^2 }{2!}+\frac{h^3(x+y)^3 }{3!}+.... $$

If you order $h^n$ of $e^{hx}e^{hy}$ and equal to $h^n$ of $e^{h(x+y)}$ , it will give you binom expansion proof.

$$\frac{(x+y)^n }{n!}=\sum_{i=0}^n \frac{x^i y^{n-i}}{i! (n-i)!} $$

$$(x+y)^n=\sum_{i=0}^n \frac{n! x^i y^{n-i}}{i! (n-i)!} $$

$$(x+y)^n=\sum_{i=0}^n {n \choose i} x^i y^{n-i} $$

Thus it is also true $$\frac{d^{n}}{dx^{n}} f(x)g(x)=\sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x)$$