Banach Contraction mapping of $\Phi(f)(x)=\int_0^x \frac{1}{1+f(t)^2}dt$ , find a fixed point. 
Let $(X,d)$ be metric space with $d(f,g)=\sup |f(x)-g(x)|$ where $X$ is the set of continuous function on $[0,1/2]$. Show $\Phi:X\rightarrow X$ $$\Phi(f)(x)=\int_0^x \frac{1}{1+f(t)^2}dt$$ has a unique fixed point $f(0)=0$.
b) Show that it satisfies $\frac{df}{dx}=\frac{1}{1+f(x)^2}$

My attempt:
I assumed $[0,1/2]$ is complete. I need to show that:
$$d(\Phi(f)(x),\Phi(g)(x))=\sup \Big|\int_0^x \frac{1}{1+f(t)^2}dt-\int_0^x \frac{1}{1+g(t)^2}dt\Big|\leq\alpha d(f,g)$$
Now,
$$\sup \Big|\int_0^x \frac{1}{1+f(t)^2}dt-\int_0^x \frac{1}{1+g(t)^2}dt\Big|=\sup \Big|\int_0^x \frac{1}{1+f(t)^2+g(t)^2}dt\Big|$$
Now ideally I need to somehow change this integral to include $d(f,g)=\sup|f(x)-g(x)|$, but I don't know how.
Then I would integrate, and get the integration constant to be $<1$, which will be my contraction constant.

A useful approach for these sorts of estimates is to use the mean value theorem.
(a) Let $g(x) = {1 \over 1+x^2}$, then $g'(x) = -{2x \over (1+x^2)^2 }$ and
$g''(x) = 2 { 3 x^2 -1 \over (1+x^2)^3 }$. It is not hard to show that
$|g'|$ has a maximum of $L={3 \sqrt{3} \over 8} <1 $ at $x = \pm { 1\over \sqrt{3}}$.
Hence $|g(x)-g(y)| \le L |x-y|$ for all $x,y$.
Then
\begin{eqnarray}
|\Phi(f_1)(x)-\Phi(f_2)(x)| &\le& \int_0^x |{1 \over 1+f_1(t)^2} - {1 \over 1+f_2(t)^2} | dt \\
&\le& \int_0^x L|f_1(t)-f_2(t)| dt \\
&\le& \int_0^x L \|f_1 - f_2\|_\infty dt \\
&\le& { 1\over 2} L \|f_1 - f_2\|_\infty
\end{eqnarray}
It follows that $\Phi$ has a unique fixed point $\hat{f}$, and
it is clear that $\hat{f}(0) = (\Phi(\hat{f})(0) = \int_0^0 {1 \over 1+\hat{f}(t)^2} dt = 0$.
(b) follows from the fundamental theorem of calculus.
