Let $$f(x)=x-\frac{1}{x-\frac{1}{....-\frac{1}{x-1}}}$$ where the fraction is iterated $2018$ times. I need to find $f'(0)$.
I'm a bit confused with this problem, I can't express it like that $f(x)=x-\frac{1}{f(x)}$ because the process is finite, how can I figure it out then?
On
Hints
$$ f(x) := \left (x-\dfrac{1}{x- \dfrac{1}{x - \dots}}\right)_{2018} $$
Then,
$$ f'(x) = 1+\left ( \dfrac{1}{x - \dfrac{1}{\dots}}\right)^2_{2017}.\left ( 1 + \left ( \dfrac 1 {x - \dfrac 1 \dots} \right)^2 \dots\right)_{2016} $$
Observe the square symbol ensures that the value inside the brackets to be positive, then
$$f'(0) = 1 + \underbrace{1^2.(1+1^2(1+1^2(1+1^2 \dots)))}_{\text{open 2018 brackets}}$$
Then, you'll get the answer 2018+1 = 2019$\blacksquare$
Let $f_n(x)$ be the function where the fraction is iterated $n-1$ times. Then $f_1(x)=x-1$ and for $n\geq 1$ $$f_{n+1}(x)=x-\frac{1}{f_n(x)}.$$ Then it is easy to verify that $f_n(0)=(-1)^n$. We show by induction that $f'_n(0)=n$. It holds for $n=1$. For the inductive step, we have that $$f'_{n+1}(x) =\frac{d}{dx}\left(x-\frac{1}{f_n(x)}\right)= 1 + \dfrac{f'_n(x)}{f_n(x)^2}$$ and for $x=0$ we find $$f'_{n+1}(0) = 1 + \dfrac{n}{((-1)^n)^2}=n+1$$ and we are done.