Leibniz's Rule

Theorem
Let $$f, g$$ be a real functions which are $n$ times differentiable at a point $$\xi$$.

Then $$D^n fg = \sum_{j=0}^n \binom n j D^j f D^{n-j} g$$ at the point $$\xi$$.

Proof
The result holds when $$n = 1$$ by an application of the Product Rule for Derivatives.

Now suppose $$D^k fg = \sum_{j=0}^k \binom k j D^j f D^{k-j} g$$.

Then $$D^{k+1} fg = \sum_{j=0}^k \binom k j \left({D^{j+1} f D^{k-j} g + D^j f D^{k-j+1} g}\right)$$, again by the Product Rule for Derivatives.

Thus (after some algebra), $$D^{k+1} fg = \sum_{j=0}^{k+1} \binom {k+1} j D^j f D^{k+1-j} g$$.

The result follows by induction.