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Let 
\begin{equation}
t=\ln(x)
\end{equation}
and define the $D$ operator such that
\begin{equation}
Dy=\frac{dy}{dt}
\end{equation}

We aim to prove for all positive integers $n$,
\begin{equation}
\frac{d^ny}{dx^n}=\frac{1}{x^n}(D)_ny
\end{equation}
\begin{equation}
\frac{d^ny}{dx^n}=\frac{1}{x^n}(D(D-1)\ldots(D-n+1))y
\end{equation}

Base case, $n=1$
\begin{equation}
\frac{dy}{dx}=\frac{dt}{dx}\frac{dy}{dt}
\end{equation}
\begin{equation}
\frac{dy}{dx}=\left(\frac{1}{x}\right)(Dy)
\end{equation}
\begin{equation}
\frac{dy}{dx}=\frac{1}{x}(D)_1y
\end{equation}
Inductive hypothesis:

Suppose that 
\begin{equation}
\frac{d^ky}{dx^k}=\frac{1}{x^k}(D)_ky
\end{equation}
\begin{equation}
\frac{d^{k+1}y}{dx^{k+1}}=\frac{d}{dx}\left[\frac{1}{x^k}\left((D)_ky\right)\right]
\end{equation}
\begin{equation}
\frac{d^{k+1}y}{dx^{k+1}}=\frac{1}{x^k}\frac{d}{dx}(D)_ky-\frac{k}{x^{k+1}}(D)_ky
\end{equation}
\begin{equation}
\frac{d^{k+1}y}{dx^{k+1}}=\frac{1}{x^k}\left(\frac{1}{x}D\right)(D)_ky-\frac{k}{x^{k+1}}(D)_ky
\end{equation}
\begin{equation}
\frac{d^{k+1}y}{dx^{k+1}}=\frac{1}{x^{k+1}}D(D)_ky-\frac{k}{x^{k+1}}(D)_ky
\end{equation}
\begin{equation}
\frac{d^{k+1}y}{dx^{k+1}}=\frac{1}{x^{k+1}}\left(D-k\right)(D)_ky
\end{equation}
\begin{equation}
\frac{d^{k+1}y}{dx^{k+1}}=\frac{1}{x^{k+1}}(D) _{k+1}y
\end{equation}
This concludes the proof.

Sat, 12 Dec 2020 07:01 GMT