Chain rule in finite calculus for ODEs (RK methods) SE.Math,
I'm reviewing for a test on numerical methods and I am having some difficulty in understanding how to take the second derivative in finite terms. I have:
$$ U(t_k)  $$ as my underlying solution and as my diff. eq
$$ \frac{dU}{dt} = F(U_k, t_k) $$
but  in order to do multistage methods, I need to take the second derive. From the Taylor's series:
$$ U(t_k) = U_k + \frac{dU}{dt}|_{t_k} \Delta t + \frac{d^2U}{dt^2}\Delta t^{2} + H.O.T$$
I know from my notes that:
$$ \frac{d^2U}{dt^2} = \frac{dF}{dt} = \frac{\delta F}{\delta U} \frac{\delta U}{\delta t} + \frac{\delta F}{\delta t} $$
but I'm lost as to how this makes sense. how can $ \frac{d^2U}{dt^2} = \frac{\delta F}{\delta t} + ...$ 
Thanks in advance! 

We know that the exact solution $U$ satisfies $\frac{d U}{d t} = F(U(t), t)$. Taking the derivative with respect to $t$ yields (in the notation of Wikipedia) 
$$ \frac{d}{d t} \left( \frac{d U}{d t} \right) = \frac{\partial F}{\partial U} \frac{\text{d} U}{\text{d} t} + \frac{\partial F}{\partial t} \frac{\text{d} t}{\text{d} t}. $$
Since $U$ is a function of a single variable ($t$), and since $\frac{\text{d} t}{\text{d} t} = 1$, this simplifies to
$$ \frac{d^2 U}{d t^2} = \frac{\partial F}{\partial U} \frac{d U}{d t} + \frac{\partial F}{\partial t}. $$
