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 (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?



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.

  • $\begingroup$ Everything is now perfectly clear to me. I deleted now irrelevant comments of mine :) $\endgroup$ – FooBar Apr 11 '15 at 18:11
  • $\begingroup$ So do I : ) ${}$ $\endgroup$ – Metta World Peace Apr 11 '15 at 18:12

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