Continued Fraction

Since the given fraction continues infinitely, we see that $f(x) = \dfrac{x}{1+ f(x)},$ or $f(x)^2 + f(x) = x.$ Implicitly differentiating both sides, we get $2f(x)f'(x) + f'(x) = 1.$ Since $f(0) = 0,$ letting $x = 0$ gives $f'(0) = \boxed{1}.$

First, note that $f(x) = \dfrac {x}{1 + f(x)}.$ We take the derivative on each side and use the Quotient Rule. We obtain: $\begin{aligned} f'(x) &= \dfrac {\left( 1 + f(x) \right) \cdot 1 - x \cdot f'(x)}{\left( 1 + f(x) \right)^2} \\&= \dfrac {1 + f(x) - x \cdot f'(x)}{\left( 1 + f(x) \right)^2}. \end{aligned}$ Substituting $x = 0$ , we have $f'(0) = \dfrac {1 + f(0) - 0 \cdot f'(0)}{\left( 1 + f(0) \right)^2}.$ Since $f(0) = 0$ , we have $f'(0) = \boxed {1}$ .

First we find an easier expression for $f(x)$ . Note that $f(x)=\frac{x}{1+f(x)}$ . Simple algebraic manipulation yields $f(x)^2+f(x)=x$ .

Next we differentiate this expression with respect to $x$ . Using the chain rule, we get $2f(x)\cdot f'(x)+1\cdot f'(x)=1$ . Again, we can manipulate this into $f'(x)=\frac{1}{2f(x)+1}$ . This is our expression for $f'(x)$ .

By this expression, we have $f'(0)=\frac{1}{2f(0)+1}$ . We have that $f(0)=\frac{0}{1+\frac{0}{1+\frac{0}{1+\cdots}}}=0$ . Thus $f'(0)=\frac{1}{2(0)+1}=\boxed{1}$ .

Notice first of all that $f(0) = \frac{0}{1 + \frac{0}{\dots}} = 0$ . We know, since the function $f(x)$ is recursive, that

$f(x) = \frac{x}{1 + f(x)} \\ (f(x))^2 + f(x) = x.$

Differentiating both sides by $x$ , we find by the Chain Rule that

$f(x) \cdot f'(x) + f'(x) = 1$

This is true for all $f(x) \ne 1$ . By plugging in $x = 0$ , we see that

$0 \cdot f'(0) + f'(0) = 1 \\ f'(0) = \boxed{1}.$