Newton Raphson method for implicit methods If $x'=\sqrt x$, to solve for implicit midpoint's method, which according to Wikipedia is,
$x_{n+1}=x_{n}+hf((x_{n+1}+x_{n})/2)$ for an autonomous system, where h is the time step, how can i use newton's method to calculate $x_{n+1}$? I did the following method,
$g(x_{n+1})=x_{n+1}-x_{n}-hf((x_{n+1}+x_{n})/2) $ and $g'(x)=\frac{g_{x+h}-g_{x}}{h}$, where $h$ is the time step. Now how can i get $x_{n+1}$? Should i use any explicit schemes or is there any other way?
 A: First consider that
$$
x_{n + 1}  + x_n  = \left( {x_{n + 1}  - x_n } \right) + 2x_n 
$$
Then consider that in "simple words" the essential of Newton-Raphson Method
is to fix an estimate of the root $x_0$, 
replace $f(x)$ with the first terms (const+first+ eventually 2nd ..) of its Taylor expansion around $x_0$
, compute therefrom the root $x=x_1$ therefrom, restart with $x_1$ ...
Of course all subject to the criteria ensuring for convergence.
So in your case we have that:
$$
\begin{gathered}
  \Delta x_n  = x_{n + 1}  - x_n  = h\,f\left( {\left( {x_{n + 1}  + x_n } \right)/2} \right) = h\,f\left( {x_n  + \Delta x_n /2} \right) =  \hfill \\
   = h\,\left( {f\left( {x_n } \right) + f'\left( {x_n } \right)\Delta x_n /2 +  \cdots } \right) \hfill \\ 
\end{gathered} 
$$
----- amended as per LutzL comment  -----
Now we should solve for $\Delta \,x_{\,n}$, keeping firm $x_n$, from
$$
\Delta \,x_{\,n}  = h\,f\left( {x_{\,n}  + \Delta \,x_{\,n} /2} \right)
$$
and for that you can use the Newton method, starting
$$
 \begin{gathered}
  {\Delta \,^{(0)} x_{\,n}} _\,  = h\,\left( {f\left( {x_{\,n} } \right) + f'\left( {x_{\,n} } \right){\Delta \,^{(0)} x_{\,n} /2}} \right) \hfill \\
{\Delta \,^{(1)} x_{\,n}} _\,  = h\,\left( {f\left( {x_{\,n}  + {\Delta ^{(0)} \,x_{\,n}} /2} \right) + f'\left( {x_{\,n}  + {\Delta ^{(0)} \,x_{\,n}} /2} \right){\Delta \,^{(1)} x_{\,n}} /2} \right)\hfill \\
\quad  \vdots 
\end{gathered} 
$$
and stopping when the result is sufficiently accurate.  
From the $\Delta \,x_{\,n}$ so found, you can get $x_{n+1}$ and reiterate the procedure
A: You have first to decide what you want to take as variable, $x_{n+1}$, $k$ in $x_{n+1}=x_n+hk$ or the midpoint $u=\frac{x_n+x_{n+1}}2$.
IMO the midpoint gives the easiest-to-read formulas. So use $x_{n+1}=2u-x_n$ to transform the implicit equation for $x_{n+1}$ into
$$
2u-x_n=x_n+hf(u)\iff 0=2u-2x_n-hf(u)
$$
The derivative of that equation is 
$$
2-hf'(u)
$$
so that you get the Newton step
$$
u_+=u-(2-hf'(u))^{-1}(2u-2x_n-hf(u))
$$
which you have to iterate until convergence in a numerical sense. Then set $x_{n+1}=2u-x_n$.
