Infinite fraction's derivative

Question

$f(x)=x+\dfrac{1}{x+\dfrac{1}{x+\dfrac{1}{x+\ldots}}}$

$f'\left(\dfrac{3}{2}\right)=?$

I tried to make equation like $y^2=xy+1$ but I can't made it clearly.

Answer 1

hint

For $ x\in [1,2]$, $ f(x)>0 $ and

$$f(x)=x+\frac{1}{f(x)} \implies$$

$$(f(x))^2-xf(x)-1=0\implies$$

$$f(x)=\frac{x+\sqrt{x^2+4}}{2}\implies$$ (take only the positive root of the quadratic)

$$f'(x)=\frac{f(x)}{\sqrt{x^2+4}}$$

You will find that $$f\left(\frac 32\right)=2\;\text{ and }\; f'\left(\frac 32\right)=\frac 45$$

Addendum :

Other expressions for $\;f’(x)\;$ are the following ones :

$f’(x)=\dfrac{f^2(x)}{f^2(x)+1}$

$f’(x)=\dfrac{f(x)}{2f(x)-x}$

Answer 2

Using implicit differentiation, $f^2-xf-1=0 $ gives $0 =2ff'-f-xf' =(2f-x)f'-f $ so $f' =\dfrac{f}{2f-x} $.

Since $f(x) =\dfrac{x+\sqrt{x^2+4}}{2} $ (need the positive root), $f(\frac32) =\dfrac{\frac32+\sqrt{\frac94+4}}{2} =\dfrac{\frac32+\sqrt{\frac{25}{4}}}{2} =\dfrac{\frac32+\frac52}{2} =2 $ so $f'(\frac32) =\dfrac{2}{4-\frac32} =\frac45 $.