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Suppose I am asked to find the slope of each x in general for $f(x)=2x$

$\Delta x$ will be a value so close to $0$ such that it doesn't matter.

Here is my approach:-

$$\frac{y_1-y_0}{x_1-x_0}=\frac{\Delta x(2)}{\Delta x}=2$$

Hence we get that for any value of $x$ in the function , the slope will be 2.

But did you notice that here we are dealing with kind of 0 in denominator.

If $\Delta x$ is so close to zero that it is doesn't matter then shouldn't $\frac{\Delta x(2)}{\Delta x}$ be indeterminate. We are not talking about evaluating the limit.$$\lim_{\Delta x\to\ 0}\frac{\Delta x(2)}{\Delta x}=2$$

Looks more agreeable to me.

How is it possible that in fractions we treat $\Delta x≠0$ and as variable but outside it we treat it like $\Delta x=0$.

This above statement makes sense. Though I would not mind doing the case of just cancelling in limits but on RHS we treat $\Delta x$ like 0 and I don't think we should do it in a limit as we are not saying what happens when we reach there. You might think that there was no $\Delta x$ obtained in RHS but I am talking in general such as in derivative of $f(x)=x^2$ would have a $\Delta x$ in RHS while evaluating the derivative.

See below for that:-

$$f'(x)=\frac{2x\Delta x + \Delta x^2}{\Delta x}=2x +\Delta x=2x$$

Did you see above while cancelling $\Delta x$ from the numerator and denominator. we treated it like a variable and while adding to 2x we treated it like 0.

It doesn't make any sense to me to use it in different ways for both purpose and especially when that we switch it(e.g. if switch $\Delta x$ to the assumption of 0 while cancelling the numerator and denominator.). We get indeterminate.

And if we do that , this way than cancelling doesn't makes any sense.( Even inside a limit of we do it that way.)

And hence this way of evaluating derivatives stop making sense.

Have I mistaken?

or

has the derivative some problem?If yes what?

Can calculus be wrong?(It looks like the probability is close to 0 for calculus being wrong and my probability of being mistake is close to 100 , if we have all mathematics correct.)

I didn't found this way of calculating derivatives much so let me tell you the source.

This method is used from - Introduction to calculus

Please avoid integrals.

Mohd Saad
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    "We are not talking about evaluating the limit." Why not? I don't fully understand your question, but it looks like you are trying to avoid talking about limits. This all makes more sense when you understand limits. – Jair Taylor Sep 13 '21 at 16:54
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    In particular, referring to "we treated it like a variable and while adding to 2x we treated it like 0. It doesn't make any sense to me to use it in different ways for both purpose..." - It's simply not true that $2x + \Delta x = 2x$, unless $\Delta x = 0$. What is true is that $\lim_{h\rightarrow 0} 2x + h = 2x$. – Jair Taylor Sep 13 '21 at 16:56
  • Unless you are talking about non-standard analysis / infinitesimals... but that is, well, non-standard. – Jair Taylor Sep 13 '21 at 16:58
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    @JairTaylor I'm looking through the linked website. I see where OPs confusion comes from. It treats a lot of this stuff very poorly. – Rushabh Mehta Sep 13 '21 at 16:59
  • @DonThousand Yeah, it looks a bit informal, which can sometimes be good but can lead to confusion. It never says something like $2x + \Delta x = 2x$, but would say that it becomes $2x$ as $\Delta x$ gets small, which is the same as taking a limit but without the word "limit". – Jair Taylor Sep 13 '21 at 17:08
  • @JairTaylor I get that physicists view this stuff this way, which is fine as long as they understand the underlying machinery, but for a beginner, this is not the way to learn. Handwaving limits only bites one in the butt later. – Rushabh Mehta Sep 13 '21 at 17:09
  • @JairTaylor I understand that in limits I can easily evaluate it. $lim_{\Delta x\to\ 0}\frac{2x\Delta x + \Delta x^2}{\Delta x}=2x+\Delta x=2x$ looks more intuitive and formal but the problem is again while cancelled we did it like a variable but after cancelling we assumed $\Delta x=0$ and added it to 2x why it is used as both 0 and variable even after being in a limit. Oh, I see as we approach the limit , we get closer and closer to 0 so we evaluated it as 2x. Am I right? If not please explain. Because again when we are evaluating it while cancelling . We use a different idea. – Mohd Saad Sep 14 '21 at 03:51
  • You are missing a "lim". It should be: $$\lim_{\Delta x \rightarrow 0} \frac{2x\Delta x + \Delta x^2}{\Delta x}= \lim_{\Delta x \rightarrow 0} (2x +\Delta x)=2x.$$ This could be just a typo but it also might be the source of confusion. You never have to assume $\Delta x = 0$, because we know that $\lim_{\Delta x \rightarrow 0} \Delta x= 0$. – Jair Taylor Sep 14 '21 at 13:01
  • Remember that when you are taking a limit, the variable won't have any specific value. eg, the $h$ in $\lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h}$ is a "dummy variable" that doesn't have any particular value but is only useful for taking a limit. – Jair Taylor Sep 14 '21 at 13:05
  • @JairTaylor Okay, I got that and it prevents confusion as well but still can you tell how are we allowed to move from one to another limit? It might be of greater help. – Mohd Saad Sep 15 '21 at 13:24
  • @JairTaylor And yes you are write missing the limit was a source of confusion. – Mohd Saad Sep 15 '21 at 13:25
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    @MohdSaad Sure, you can say $$\lim\limits_{\Delta x\to 0}\frac{2x\Delta x+\Delta x^2}{\Delta x}=\lim\limits_{\Delta x\to 0}2x+\Delta x$$ since $\frac{2x\Delta x+\Delta x^2}{\Delta x} = 2x + \Delta x$ for all $\Delta x \neq 0$. (When taking limits the value at the point doesn't matter, only the values near the point.) In other words, you're just simplifying the expression here; you haven't evaluated the limit yet. – Jair Taylor Sep 15 '21 at 15:43
  • @JairTaylor Okay, Now I see what we are doing exactly. – Mohd Saad Sep 16 '21 at 06:15

1 Answers1

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Let's take a look at one of the equations you provide:$$f'(x)=\frac{2x\Delta x + \Delta x^2}{\Delta x}=2x +\Delta x=2x$$As you suggest, this is not valid. Either $\Delta x=0$, or not. Either way, we have an issue here.

So let's do this a bit better. Let $f(x)=x^2$. How do we compute $f'(x)$? Well, by the definition of the derivative: $$f'(x)=\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$Note what this means. When we take a limit, we say that we consider the function as the limiting variable (in this case $\Delta x$) approaches, but does not equal, the limiting value (in this case $0$). Evaluating, we get $$\lim\limits_{\Delta x\to 0}\frac{(x+\Delta x)^2-x^2}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{2x\Delta x+\Delta x^2}{\Delta x}=\lim\limits_{\Delta x\to 0}2x+\Delta x=2x$$Since $\Delta x\neq0$, we could move from the second limit to the third. Then, we take the limit. The closer $\Delta x$ gets to $0$, the closer $2x+\Delta x$ gets to $2x$. So the limit is $2x$.

If limits are confusing, it's worth it to take the time to understand them. Khan Academy is a fine source

Rushabh Mehta
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  • @Mohd Saad: In addition to what Don Thousand said, see this answer by Paramanand Singh. – Dave L. Renfro Sep 13 '21 at 17:35
  • @DonThousand That is what exactly I , thought. What I have concluded is:-In the last limit we don't have anything to factor out or cancel but we see as $\Delta x$ gets closer to 0 the last limit gets closer to 2x. But limits don't tell what exactly is going there so how can we say that 2x is the sure derivative. – Mohd Saad Sep 14 '21 at 03:56
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    @MohdSaad A limit is precisely that: a statement saying we can get arbitrarily close to $2x$. The point is, the derivative is defined in terms of a limit, so that means that the derivative is exactly $2x$. To put it in other words, to say that the derivative is $2x$ is to say that by making $\Delta x$ small, we can get the fraction $\frac{f(x+\Delta x)-f(x)}{\Delta x}$ arbitrarily close to $2x$. (This is actually not exactly right, but that's not too important.) – Rushabh Mehta Sep 14 '21 at 11:47
  • @DonThousand So can I interpret it like that the limit is so close that it is exact. – Mohd Saad Sep 15 '21 at 11:41
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    @MohdSaad A limit is as close as you want it to be. That's essentially what it means (although learning the technical definition is important, too) – Rushabh Mehta Sep 15 '21 at 12:30
  • @DonThousand Oh, I just did learnt the epsilon-delta definition of limits only from the website you linked. It happens all started to make sense but I think if you can explain in a bit more details *why we can move from one the first limit to second and from second to thousand?* – Mohd Saad Sep 15 '21 at 13:23
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    The first limit to the second is just algebraic manipulation. The second to the third is factoring out a $\frac{\Delta x}{\Delta x}$ term from the product, but since $\Delta x\neq0$, this fraction is just $1$, so we can discard it. – Rushabh Mehta Sep 15 '21 at 13:27
  • @DonThousand Okay, I got that all we are doing is manipulation to evaluate our limit , right? – Mohd Saad Sep 16 '21 at 05:00
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    @MohdSaad Correct – Rushabh Mehta Sep 16 '21 at 11:54