<p>$$\require{cancel}$$</p>

<p>Notice that if $\dot x_t \neq 0$, then $\Delta \dot x_t \to 0 \implies \Delta \to 0$</p>

<p>$$ \lim_{\Delta\to 0} \frac{f(x_{t+\Delta}) - f(x_t)}{\Delta} = \cancelto{f'(x_t)}{\lim_{\Delta \dot x\to 0} \frac{f(x_t + \Delta \dot x_t) - f(x_t)}{\Delta \dot x_t} \dot x_t}= f'(x_t)\dot x_t$$</p>

<p>If $\dot x_t = 0$, then $f(x_t)$ is a constant function.</p>

CC BY-SA 3.0
2015-04-11T18:11:26.260Z
Everything is now perfectly clear to me. I deleted now irrelevant comments of mine :)
FooBar
CC BY-SA 3.0
2015-04-11T18:12:28.873Z
So do I : ) ${}$
Metta World Peace
CC BY-SA 3.0
2015-04-11T17:51:31.490Z
2015-04-11T18:07:20.977Z
2015-04-11T18:07:20.977Z
Metta World Peace
<p>I know one form of a derivative, which is</p>

<p>$$ f'(x) = \lim_{\Delta\to 0} \frac{f(x+\Delta) - f(x)}{\Delta}$$</p>

<p>Let $x_{t+\Delta} = x_t + \Delta \dot x_t$. <a href="http://www.princeton.edu/~moll/HACT.pdf" rel="nofollow">Achdou and coauthors</a> (end of appendix A1) claim that</p>

<p>$$ \lim_{\Delta\to 0} \frac{f(x_{t+\Delta}) - f(x_t)}{\Delta} = \lim_{\Delta\to 0} \frac{f(x_t + \Delta \dot x_t) - f(x_t)}{\Delta} = f'(x_t)\dot x_t$$</p>

<p>I understand the first equality, that's merely the substitution. I can't wrap my head around the second one. How exactly does one show it? </p>

<mathematical-economics>
From Discrete to Continuous Time
FooBar
# From Discrete to Continuous Time

(Asked by: username_0 on 2015-04-11)

I know one form of a derivative, which is

$$ f'(x) = \lim\_{\Delta\to 0} \frac{f(x+\Delta) - f(x)}{\Delta}$$

Let $x\_{t+\Delta} = x\_t + \Delta \dot x\_t$. [Achdou and coauthors](http://www.princeton.edu/~moll/HACT.pdf) (end of appendix A1) claim that

$$ \lim\_{\Delta\to 0} \frac{f(x\_{t+\Delta}) - f(x\_t)}{\Delta} = \lim\_{\Delta\to 0} \frac{f(x\_t + \Delta \dot x\_t) - f(x\_t)}{\Delta} = f'(x\_t)\dot x\_t$$

I understand the first equality, that's merely the substitution. I can't wrap my head around the second one. How exactly does one show it?

**Tags:** mathematical-economics


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## 1 answers

### Answer by username_1 on 2015-04-11. Score: 2 | Accepted answer

$$\require{cancel}$$

Notice that if $\dot x\_t \neq 0$, then $\Delta \dot x\_t \to 0 \implies \Delta \to 0$

$$ \lim\_{\Delta\to 0} \frac{f(x\_{t+\Delta}) - f(x\_t)}{\Delta} = \cancelto{f'(x\_t)}{\lim\_{\Delta \dot x\to 0} \frac{f(x\_t + \Delta \dot x\_t) - f(x\_t)}{\Delta \dot x\_t} \dot x\_t}= f'(x\_t)\dot x\_t$$

If $\dot x\_t = 0$, then $f(x\_t)$ is a constant function.

> **Comments:**
> **username_0**: Everything is now perfectly clear to me. I deleted now irrelevant comments of mine :)  
> **username_1**: So do I : ) ${}$  

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