Show that $\frac{d^{n}}{dx^{n}}\left[ x^{n-1}f\left(\frac{1}{x}\right)\right]=\frac{(-1)^{n}}{x^{n+1}}f^{(n)}\left(\frac{1}{x}\right)$ How can I prove that
$$\frac{d^{n}}{dx^{n}}\left[ x^{n-1}f\left(\frac{1}{x}\right)\right]=\frac{(-1)^{n}}{x^{n+1}}f^{(n)}\left(\frac{1}{x}\right)$$
without using mathematical induction? For any $n\in\mathbb{N}$.
 A: Ok, my first answer was -- admittedly -- not that helpful so here's another one:


*

*You can try to prove the result for a smaller class of functions $f$, say analytical functions.
Because the map $$T : f \mapsto \frac{d^n}{dx^n}[x^{n-1}f(1/x)] - \frac{(-1)^n}{x^{n+1}}f^{(n)}(1/x) $$
is linear in $f$, we may even try to prove the result for monomials $f(x) = x^N$ first.
$$ \frac{d^n}{dx^n}[x^{n-1}f(1/x)]
    = \frac{d^n}{dx^n}x^{n-1-N}
    = (n-1-N)^{\underline{n}}x^{-1-N}
   $$
Here I used the notation: $a^{\underline{b}}$ is the $b$-th falling factorial of $a$.
Further we have
$$ \frac{(-1)^n}{x^{n+1}}f^{(n)}(1/x)
    = (-1)^nx^{-(n+1)}\frac{d^n}{ds^n}|_{s=1/x}s^{N}
    = (-1)^nx^{-(n+1)}N^{\underline{n}}x^{-(N-n)}
    = (n-1-N)^{\underline{n}}x^{-1-N}
   $$
Hence $T(f) = 0$ for monomials $f$.

*By linearity $T$ also vanishes on polynomials.

*Without loss we may assume that $f(1/x)$ is defined on a bounded interval $I=[a,1/a]$ for $a\in (0,1)$, so $T : C^{n}(I) \to C^{0}(I)$. $C^{n}(I)$ is equipped with the norm $\|f\|_{C^n} := \max\{\|f\|_\infty, ..., \|f^{(n)}\|_\infty \}$. $C^0(I)$ is equipped with the uniform norm. We need to show that $T$ is continuous.

*Now it remains to show that the polynomials on $I$ are dense in $C^n(I)$, which has been shown here using Jackson's theorem.
