<p>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}</p>

CC BY-SA 3.0
2015-09-22T07:17:45.523Z
2015-09-22T07:17:45.523Z
Oliv
<p>I'm trying to derive an HJB from a discrete time setting. At some point, I am left with</p>

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

<p>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.</p>

<p>How do I proceed with the expression above?</p>

CC BY-SA 3.0
2015-09-21T14:17:29.443Z
If I understood well your question, you try to show where the expression comes from. It is the basic difference quotient. The logic of total derivative can be understood by this one. Is it what you are looking for ?
optimal control
CC BY-SA 3.0
2015-09-21T14:40:05.107Z
@optimalcontrol "The logic of total derivative can be understood by this one.' which is what I am looking for. I'd like to see the derivation of the total derivative, starting from the above expression.
FooBar
CC BY-SA 3.0
2015-09-21T22:12:35.910Z
I have showed the most part on my papers but I will try to write down on LateX tomorrow when I have time. 
Also, altough the content is about economics, I think your question fits better for mathematics.stackexchange.com because what you are looking for as an explanation is purely mathematics.
optimal control
<mathematical-economics><continuous-time>
From Discrete to Continuous time: Total Differential
FooBar
# From Discrete to Continuous time: Total Differential

(Asked by: username_0 on 2015-09-21)

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?

**Tags:** mathematical-economics, continuous-time

> **Comments:**
> **username_1**: If I understood well your question, you try to show where the expression comes from. It is the basic difference quotient. The logic of total derivative can be understood by this one. Is it what you are looking for ?  
> **username_0**: @optimalcontrol "The logic of total derivative can be understood by this one.' which is what I am looking for. I'd like to see the derivation of the total derivative, starting from the above expression.  
> **username_1**: I have showed the most part on my papers but I will try to write down on LateX tomorrow when I have time.  Also, altough the content is about economics, I think your question fits better for mathematics.stackexchange.com because what you are looking for as an explanation is purely mathematics.  

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## 1 answers

### Answer by username_2 on 2015-09-22. Score: 6 | Accepted answer

You can separate your function in three terms by writing
\begin{align}
& v(c\_{t+\Delta},u\_{t+\Delta},t+\Delta)-v(c\_t,u\_t,t) = \\
& v(c\_{t+\Delta},u\_{t+\Delta},t+\Delta)-v(c\_t,u\_{t+\Delta},t+\Delta) \\
+ & v(c\_{t},u\_{t+\Delta},t+\Delta)-v(c\_t,u\_t,t+\Delta) \\
+ & 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}


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