From Discrete to Continuous time: Total Differential I'm trying to derive an HJB from a discrete time setting. At some point, I am left with

$$ \lim_{\Delta\to 0} \frac{v(c_{t+\Delta}, u_{t+\Delta}, t+\Delta) - v(c_{t}, u_{t}, t)}{\Delta}$$

and am not sure what to do. If $\Delta$ was only in one argument, this would be a partial differential. My hunch is that this is the total derivative in $t$, but I dont know how to show that.

How do I proceed with the expression above?
You can separate your function in three terms by writing
\begin{align}
&amp; v(c_{t+\Delta},u_{t+\Delta},t+\Delta)-v(c_t,u_t,t) = \\
&amp; v(c_{t+\Delta},u_{t+\Delta},t+\Delta)-v(c_t,u_{t+\Delta},t+\Delta) \\
+ &amp; v(c_{t},u_{t+\Delta},t+\Delta)-v(c_t,u_t,t+\Delta) \\
+ &amp; v(c_t,u_t,t+\Delta)-v(c_t,u_t,t)
\end{align}
When you divide by $\Delta$ and take the limit $\Delta \rightarrow 0$, the first expression converges to $\dfrac{\partial v}{\partial c} \dfrac{dc}{dt}$, the second expression to $\dfrac{\partial v}{\partial u} \dfrac{du}{dt}$, and the third expression to $\dfrac{\partial v}{\partial t}$. Therefore your total derivative equals
\begin{equation}
\dfrac{\partial v}{\partial c} \dfrac{dc}{dt} + \dfrac{\partial v}{\partial u} \dfrac{du}{dt} + \dfrac{\partial v}{\partial t}
\end{equation}