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:
$$ \frac{d^n}{dx^n}f(x)g(x)= \frac{d^{n}f(x)}{x^n}g(x)+f(x)\frac{d^{n}g(x)}{x^n}?$$
On
No, you can not here is a simple counterexample take
$$f(x)=x \hspace{10pt} g(x)=e^x$$
$$\frac{d^2(f(x)g(x))}{dx^2}=2e^x+xe^x$$ while $$\frac{d^2x}{dx^2} e^x + x \frac{d^2(e^x)}{dx^2}=0+xe^x=xe^x$$
$$\frac{d^2(xe^x)}{dx^2}\ne \frac{d^2x}{dx^2} e^x + x \frac{d^2(e^x)}{dx^2}$$
On
In comments, you asked why it never holds. It sometimes holds (just not always). Consider the two following (simple) examples:
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.
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).
The generalisation you are seeking is Leibniz' formula: $$\frac{d^n}{dx^n}(f(x)g(x)) =\sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac{d^k f(x)}{dx^k}\frac{d^{n-k} g(x)}{dx^{n-k}}.$$