Question: If $f(x) = \cfrac{x}{x + \cfrac{x}{x + \cfrac{x}{x + \vdots}}}$, find $f'(x)$.
Possible Answer:
After searching all over the place I’ve given up and resorted to posting this, I know this is a very simple problem for most of you on here but I was wondering if you could check my answer- found this question somewhere on this website yet they didn’t post the answer.
Thanks
George
Since we have that \begin{align} f(x) = \frac{x}{x+f(x)} \end{align} then it follows \begin{align} f'(x) = \frac{x+f(x)-x-xf'(x)}{(x+f(x))^2} \ \ \implies f'(x)\left(1+\frac{x}{(x+f(x))^2} \right) = \frac{f(x)}{(x+f(x))^2}. \end{align} Hence it follows \begin{align} f'(x)=\frac{f(x)}{(x+f(x))^2+x}. \end{align}
Edit: Observe that \begin{align} f'(x) =&\ \frac{f(x)}{x^2+2xf(x)+f(x)^2+x} = \frac{f(x)}{x^2+xf(x)+2x} = \frac{f(x)+x-x}{x^2+xf(x)+2x}\\ =&\ \frac{1-\frac{x}{x+f(x)}}{x\frac{x+f(x)}{x+f(x)}+2\frac{x}{x+f(x)}} = \frac{1-f(x)}{x+2f(x)}. \end{align}
I have a somewhat different answer. Starting from $$\begin{align} f(x) = \frac{x}{x+f(x)} \end{align}$$ we have $$xf(x)+f^2(x)=x\\ xf'(x)+f(x)+2f(x)f'(x)=1\implies\\ f'(x)=\frac{1-f(x)}{x+2f(x)}$$
If this solution and Jacky Chong's are both correct (and his certainly looks good to me) we ought to be able to equate them and solve for $f(x).$
EDIT As pointed out in KingW3's answer, instead of equating the derivatives, it's easier to just just solve $$f^2(x)+xf(x)-x=0$$ for $f(x)$ in terms of $x$. This gives $$f(x)=\frac{-x\pm\sqrt{x^2+4x}}{2}$$ From the well-known continued fraction for the golden section, we know that $$f(1)=\frac{-1+\sqrt 5}{2},$$ so we must pick the plus sign, and $$\boxed{f(x)=\frac{-x+\sqrt{x^2+4x}}{2}}$$
From $xf(x) +f^2(x)=x$ putting $t=f(x)$ we get a quadratic in $t$ and $t_{1,2}=\frac{-x\pm\sqrt{x^2-4x}}2$. Since you have two different functions you get two different derivatives depending on what $f(x)$ you choose. In this particular case $f(x)$ is not uniquely defined so finding it's derivative is impossible.
