derivation of order 3 method for differential equations I am stuck with a big problem.
Trying to understand the proof that the numerical method of solving differential equation

$x_{i+1} = x_i + \tau_iF(t_i+\frac{\tau_i}{2}, x_{i+\frac{1}{2}})$
$x_{i+\frac{1}{2}} = x_i + \frac{\tau}{2}F(t_i,x_i)$

which seems to be the Cauchy-Euler's method.
I am stuck with the Taylor approximation of

$x(t+\tau) = x(t)+\tau x'(t) + \frac{1}{2} x''(t) + O(\tau^3)$.

Here it is OK for me.
I also understand, why $x'(t)=F(t,x(t))$.
But why

$x''(t) = \frac{d}{dt}F(t,x(t)) = \partial_tF(t,x(t)) + D_xF(t,x(t))F(t,x(t))$ ???

I am a little frustrated about this.
Normally I am more into computer science and some mathematical concepts can be missing, so sorry for possible trivial or illposed question, thanks for patience :)
 A: Consider $F$ as a function of $t$ and $x$: then we have 
$$
\begin{split}
x''(t)&=\lim_{h\to 0}\frac{x'(t+h)-x'(t)}{h}=\lim_{h\to 0} \frac{F(t+h,x(t+h))-F(t,x(t))}{h}\\
&=\lim_{h\to 0} \frac{F(t+h,x(t+h))- F(t,x(t+h))+F(t,x(t+h))-F(t,x(t))}{h}\\
&=\lim_{h\to 0} \frac{F(t+h,x(t+h))- F(t,x(t+h))}{h}+\lim_{h\to 0}\frac{F(t,x(t+h))-F(t,x(t))}{h}\\
&\triangleq \partial_tF(t,x(t))+D_xF(t,x(t))\cdot x'(t)=\partial_tF(t,x(t))+D_xF(t,x(t))\cdot F(t,x(t)).
\end{split}
$$
The last term in the last line of the above equality is obtained by using the chain rule.
A: This is really a matter of applying the multivariable chain rule, and being clear on what your notation means. Hopefully you know that $\partial_tF$ means the function you get when you differentiate $F$ with respect to its first argument, and $(\partial_tF)(t,x(t))$ means the value of this function when you evaluate it at the point $(t,x(t))$, and likewise for the other symbols.
So, if you directly apply the chain rule, you get the following:
\begin{equation}
x''(t) = (\partial_tF)(t,x(t)) + (D_xF)(t,x(t)) \cdot x'(t)
\end{equation}
But notice that we also have $x'(t) = F(t,x(t))$. So, substituting this, we get
\begin{equation}
x''(t) = (\partial_tF)(t,x(t)) + (D_xF)(t,x(t)) \cdot F(t,x(t))
\end{equation}
