TITLE: Why does $n$-time differentiation of product have the same structure as raising sum to $n$th power?
QUESTION [17 upvotes]: A formula for differentiating a product is well known:
$$(ab)'=a'b+ab'.$$
At first sight it doesn't resemble anything interesting. But what if we differentiate twice? We'll get
$$(ab)''=a''b+2a'b'+ab''.$$
Now this does resemble something:
$$(a+b)^2=a^2b^0+2a^1b^1+a^0b^2.$$
Of course, going to the next level now gives us the expected result:
$$(ab)'''=a'''b+3a''b'+3a'b''+ab''',$$
in full accordance to known binomial expansion:
$$(a+b)^3=a^3b^0+3a^2b^1+3a^1b^2+a^0b^3.$$
I know that binomial expansion stems from distributive law of multiplication over addition. It's easy because integer powers can be interpreted as iterated multiplication. But I can't seem to see any simiar distributive law of differentiation over product.
Observing this, I wonder: how to explain this striking similarity between seemingly unrelated things? Can this be explained by some generalized form of distributive law?

REPLY [2 votes]: I hope this site is ready for some funky notation, because here it comes (but maybe not immediately):
Firstly, let's denote $a'$ as $Da$ instead and $a''$ as $D^2 a$ and so on, where $D$ is the map taking $f$ to its derivative.  Then, the expression
$$Da\,b+a\,Db$$
looks to me like the sum of two expressions, acting on the pair $(a,b)$ - the first expression, $Da\,b$ takes the derivative of the first and multiplies it by the second. The second expression does that, in reverse, more or less.
To formalize this a bit (this is where the funky notation comes - the curly braces do not refer to sets), let's denote that first expression - where we map $(a,b)$ to $Da\,b$ - as $\{1,D\}$ and the second expression as $\{D,1\}$. Then, we are basically saying that the derivative of $ab$ equals the application of $$\{1,D\}+\{D,1\}$$
to $(a,b)$.
However, a notable property of this notation is that if we could write $ab$ as the application of $\{1,1\}$ to $(a,b)$ - that is, we multiply $a$ and $b$ by $1$, then multiply them together. So, we're basically saying that, in a derivative, we have
$$D(\{1,1\}\cdot (a,b))=(\{1,D\}+\{D,1\})\cdot(a,b)$$
where $\{x,y\}\cdot (a,b)$ is the application of $\{x,y\}$ to $(a,b)$. However, if we apply this multiple times we get that
$$D^2(\{1,1\}\cdot (a,b))=D((\{1,D\}+\{D,1\})\cdot(a,b))$$
which, since the derivative is linear, it should, in some sense, commute with the curly brace operators, which would give
$$D^2(\{1,1\}\cdot (a,b))=(\{1,D\}+\{D,1\})D(\{1,1\}\cdot(a,b))$$
$$D^2(\{1,1\}\cdot (a,b))=(\{1,D\}+\{D,1\})^2\cdot(a,b)$$
$$D^2(\{1,1\}\cdot (a,b))=(\{1,D^2\}+2\{D,D\}+\{D^2,1\})\cdot(a,b).$$
where we'd expand the right side as $ab''+2a'b'+2a''b$. (Note that we're taking the product of curly-brace operators to be $\{x_1,y_1\}\{x_2,y_2\}=\{x_1x_2,y_1y_2\}$)
More generally, we will get
$$D^n(\{1,1\}\cdot (a,b))=(\{1,D\}+\{D,1\})^n\cdot(a,b)$$
where we clearly have an exponent over a sum, which yields the relation. All of the above could be formalized, if so desired, but given that the question seems to ask for intuition, that's likely unnecessary.
(Though, ultimately, this is basically the same as Danijel's answer, which was posted before I finished this, just expressed in different terms)