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Suppose an example where you are told to find the derivative of any function say, $x^2$.

$$\frac{\Delta y}{\Delta x} = \frac{f(x + \Delta x ) - f(x)}{\Delta x}$$

Using $f(x) = x^2$, and upon solving you're left with:

$$\frac{\Delta y}{\Delta x} = 2x + \Delta x$$

And to get the actual result, I was told to substitute $\Delta x = 0$ and that is my question. How does it make sense to do so?

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cpx
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    Well, try not doing so and see what happens. Calculus of finite differences has its applications, after all. – Ivan Neretin Sep 25 '15 at 07:21
  • PS: I have no idea how to make the math fonts size larger. – cpx Sep 25 '15 at 07:21
  • If you want a formula on a line of its own, enclose it with $$ instead of $. (That also makes some things, e.g. fractions, larger.) – mrf Sep 25 '15 at 07:23
  • Use \displaystyle or \dfrac, I guess. $\frac{df}{dx}$ will be $\dfrac{df}{dx}$. – Ivan Neretin Sep 25 '15 at 07:24
  • Do you know what a limit is? –  Sep 25 '15 at 07:25
  • An abstract infinitesimal is an arbitrarily or indefinitely small value. It can be made immeasurable regardless of how you define that. In other words - just because you can't put a value on it, it doesn't mean it's zero. But you can neglect it anyway. Relevant. –  Sep 29 '16 at 22:58

2 Answers2

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Because the derivative is defined as

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

So, for example, if the function $\Delta x \mapsto \frac{f(x+\Delta x) - f(x)}{\Delta x}$ is continuous around $\Delta x = 0$, then you can calculate the limit by simply plugging in $\Delta x=0$.

5xum
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  • Exactly. When using the definition to calculate derivatives, you just need to work on the expression first so it doesn't become "$\frac00$" when you carry out the substitution $\Delta x \to 0$. – Arthur Sep 25 '15 at 07:22
  • I saw but I want to know why does it become 0 and how does it make sense in case of slope? – cpx Sep 25 '15 at 07:22
  • @cpx If $g(x)$ is continuous at $0$, then $\lim_{x\to 0} g(x) = g(0)$. – 5xum Sep 25 '15 at 07:25
  • Sorry I don't know what "continuous" is. I updated the question with a diagram. Can you say that in layman terms perhaps? – cpx Sep 25 '15 at 07:32
  • @cpx If you don't know what continuous is, you have no business calculating derivatives. Sorry, but that's the blunt truth. You have to learn to walk before you can run. – 5xum Sep 25 '15 at 07:33
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    With the sine at $x=0$, we have $f'(0)=\frac{\sin(\Delta x)}{\Delta x}$ and obviously this is continuous around $\Delta x=0$, but you certainly cannot plug in $\Delta x=0$, can you? You meant something slightly different, I guess. – egreg Sep 25 '15 at 07:47
  • @egreg You cannot, because the expression is not continuous around $0$, because it is not defined at $0$. – 5xum Sep 25 '15 at 07:48
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    @5xum *No* expression of the form $\frac{f(x+\Delta x)-f(x)}{\Delta x}$ is defined at $\Delta x=0$. – egreg Sep 25 '15 at 07:49
  • @egreg No, but the expression $2x + \Delta x$ IS. – 5xum Sep 25 '15 at 07:57
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    @5xum This is exactly why people get confused when learning derivatives. – egreg Sep 25 '15 at 08:12
  • @egreg Yeah. I would improve my answer, but I don't have the motivation. The OP clearly says he does not know what continuous means. As far as I am concerned, it makes no sense to explain derivatives to someone who knows nothing about continuity. – 5xum Sep 25 '15 at 08:14
  • @5xum: Sorry if you feel discouraged but I just picked up that simple little concept while I was looking and your answer would be much appreciated! – cpx Sep 25 '15 at 08:17
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The finite-difference expression $g(x,\Delta x)=\dfrac{f(x+\Delta x)-f(x)}{\Delta x}$ is depending on two parameters, $x$ and $\Delta x$. The first indicates a "location", and the second a "scale".

Working with these two parameters is much more complicated than a single one, and calculus would certainly have made much slower progress without the possibility of getting rid of the scale.

It turns out that when the functions are smooth (local fluctuations decreasing with scale), the finite difference stabilizes to a constant value when you reduce $\Delta x$, and this constant value is representative of the local behavior of the function.

Hence, by convention you extend $g(x,\Delta x)$ to $g(x,0)$ by assigning this constant value.

Actually, the nuance between finite-difference and derivative (infinitesimal difference if you like) indeed appears in your example

$$\frac{\Delta f}{\Delta x}=2x+\Delta x,$$ which is more complicated than $$\frac{df}{dx}=2x.$$ The difference may seem futile, but you quickly see that it is not:

$$\frac{\cos(x+\Delta x)-\cos(x)}{\Delta x}=\frac{\cos(x)\cos(\Delta x)-\sin(x)\sin(\Delta x)-\cos(x)}{\Delta x}$$ does not simplify, whereas $$\cos'(x)=-\sin(x).$$