Can you distribute the nth derivative via the product rule? I just want to check a particular step I am trying out. Normally I would prove an nth derivative formula via induction, but I am wondering if there is an easier way to potentially make this work. Is it true for some possible manipulation of this formula for analytic functions f(x) and g(x) for $ x \in {\Bbb R}$ that:

$$ \frac{d^n}{dx^n}f(x)g(x)= \frac{d^{n}f(x)}{x^n}g(x)+f(x)\frac{d^{n}g(x)}{x^n}?$$
The generalisation you are seeking is Leibniz' formula:
$$\frac{d^n}{dx^n}(f(x)g(x))
=\sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac{d^k f(x)}{dx^k}\frac{d^{n-k} g(x)}{dx^{n-k}}.$$
In comments, you asked why it never holds.  It sometimes holds (just not always).  Consider the two following (simple) examples:


$f(x)$ could be anything, and $g(x)=1$.


In this case, $\frac{d^{n}}{dx^n} f(x)g(x)=\frac{d^{n}}{dx^n} f(x)\cdot 1 = \frac{d^{n}f(x)}{dx^{n}}(1)+f(x)\cdot 0$ and the formula works.


Suppose that $f(x)=g(x)=x^2$ so that $f(x)g(x)=x^4$ and take $n=4$.


By easy computation $\frac{d^4}{dx^4} x^4 = 24$.

But, using your formula you would get:

$\frac{d^4 (x^2)}{dx^4} x^2 + x^2 \frac{d^4 (x^2)}{dx^4} = 0\cdot x^2 + x^2 \cdot 0 = 0$ which disagrees with what you know the derivative should be (24).