Let $f: \mathbb R^+ \to \mathbb R $
$$f=\mathop{\vcenter{\LARGE\mathrm K}}\limits_{j=1}^{+\infty}\frac{x}{x^j}=\cfrac{x}{x^1+\cfrac{x}{x^2+\cfrac{x}{x^3+\ddots}}}$$
I saw this problem in a math challenges magazine, I have no idea how this would be solved but thought it seemed really interesting.
We have: $$ f(x) = \cfrac{x}{x^1+\cfrac{x}{x^2+\cfrac{}{}}}=\cfrac{1}{1+\cfrac{1/x^2}{1+\cfrac{1/x^4}{1+\ldots}}}=x^{2/5}\,R\left(\frac{1}{x^2}\right)\tag{1}$$ where $R(q)$ is the Rogers-Ramanujan continued fraction, so: $$ f(x) = \prod_{k\geq 0}\frac{(1-\frac{1}{x^{10k+2}})(1-\frac{1}{x^{10k+8}})}{(1-\frac{1}{x^{10k+4}})(1-\frac{1}{x^{10k+6}})}\tag{2}$$ and $f'(x)$ can be computed as $f(x)\cdot\frac{d}{dx}\log f(x)$, where the logarithmic derivative of $f(x)$ is given by a nice series. It is also interesting to recall that by setting $u=R(q)$ and $v=R(q^2)$ we have: $$ v-u^2 = (v+u^2) uv^2. \tag{3}$$ Since $R(q)$ is an analytic function over the unit ball, $f(x)$ is an analytic function over $(1,+\infty)$.