I computed the first few derivatives in the sequence $\frac{1}{x+\frac{1}{x}}, \frac{1}{x+\frac{1}{x+\frac{1}{x}}} \dots$ and eventually lost feeling in my fingers and also realized I was seeing nothing meaningful. Let
$$f(x,t) = \overbrace{\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{\ddots}}}}^{x\text{ appearing }t \text{ times.}}$$
My question: does there exist a closed-form of the derivative of an infinite continued fraction, at least for positive arguments? That is, does $g(x)$ such that: $$g(x)=\lim_{t \to \infty}\frac{d}{dx}f(x,t)$$ Have a closed-form expression?
Let $y:=\frac{1}{x+\frac{1}{x+\cdots}}=\frac{1}{x+y}$ so$$y^2+xy-1=0\implies y=\frac{-x\pm\sqrt{x^2+4}}{2}.$$If $x>0$, $y>0$ so$$y=\frac{-x+\sqrt{x^2+4}}{2}\implies\frac{dy}{dx}=\frac{-1+\frac{x}{\sqrt{x^2+4}}}{2}.$$Your $f(x,\,t)$ are odd for each integer $t\ge0$, so if we are to extend this to $x\le0$ we need$$y=\frac{-x+\operatorname{sgn}x\cdot\sqrt{x^2+4}}{2},\,y^\prime=\frac{-1+\frac{|x|}{\sqrt{x^2+4}}}{2}.$$