This is an exercise from Morris Kline's "Calculus: An Intuitive and Physical Approach":

What is $\lim\limits_{\Delta x \to 0} \frac{f(x + 2 \Delta x) - f(x)}{\Delta x}$?
Suggestion: Let $2 \Delta x = t$

Following the hint, we have

\begin{align} \lim\limits_{\Delta x \to 0} \frac{f(x + 2 \Delta x)-f(x)}{\Delta x} &=\lim\limits_{t \to 0} \frac{f(x + t)-f(x)}{t/2} \\ &= \lim\limits_{t \to 0} 2 \left ( \frac{f(x + t)-f(x)}{t} \right) \\ &= 2 \left ( \lim\limits_{t \to 0} \frac{f(x + t)-f(x)}{t} \right) \\ &= 2 f'(x) \end{align}

I'm not fully understanding why this is true. Wouldn't it depend on the function we are differentiating? How can we be sure that increasing the change in $x$ will increase the instantaneous rate of change of in $f(x)$?


Your calculation is fine. For an intuition, we are considering the variation of the variable for the function $2\Delta x$ twice the variation of the variable $\Delta x$ therefore the result we obtain is twice the derivative at that point.

More in general

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

  • $\begingroup$ So basically the instantaneous rate of change increases by a factor of $n$ since we increased how much the function varies by a factor of $n$ while keeping the interval of variation the same ($\Delta x$)? $\endgroup$ – Iyeeke Sep 9 '20 at 20:53
  • $\begingroup$ @Iyeeke Yes exactly that's the reason why we obtain $nf'(x)$. $\endgroup$ – user Sep 9 '20 at 20:54
  • $\begingroup$ I see, thank you! $\endgroup$ – Iyeeke Sep 9 '20 at 20:58

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