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In calculus we are taught that the derivative of function $y$ with respect to $x$ is defined as "the quantity which $\dfrac{\bigtriangleup{y}}{\bigtriangleup{x}}$ tends to when $\bigtriangleup{x}$ tends to zero"

Can we also define it as "the ratio of $\dfrac{\bigtriangleup{y}}{\bigtriangleup{x}}$ when $\bigtriangleup{x}$ is a quantity very close to zero"?

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    $\begingroup$ No, at least for standard analysis. First, ‘very close’ is not really mathematically defined. Second, any value of this ratio has no reason to be equal to the limit. Informally speaking, it will be ‘very close’, but not equal in general. $\endgroup$ – Bernard Mar 26 '17 at 15:05
  • $\begingroup$ No.That's only an approximation of the derivative. $\endgroup$ – Thomas Andrews Mar 26 '17 at 15:05
  • $\begingroup$ Consider the difference between a line tangent to a circle and a line that hits the circle at two points very close together. The latter might have a slope that is close to the slope of the tangent, but it won't be the same line. $\endgroup$ – Thomas Andrews Mar 26 '17 at 15:07
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    $\begingroup$ Well, I consider both to be very informal but, no, If $\Delta x$ is a value very close to zero then $\frac {\Delta y}{\Delta x}$ well be a value when x is a single number when x is close to be not equal to 0 and that is different then the ultimate limit of $\frac {\Delta y}{\Delta x}$ tends to but might not ever be. $\endgroup$ – fleablood Mar 26 '17 at 16:21
  • $\begingroup$ Please replace \bigtriangleup by \Delta. $\endgroup$ – Did Apr 25 '17 at 7:19
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Let $y = x^2$. Then $\dfrac{dy}{dx} = 2x$ for the usual reasons

$\lim_{\Delta x \rightarrow 0} \dfrac {\Delta y}{\Delta x} = \lim \dfrac {[x^2 + 2x\Delta x + (\Delta x)^2] -[x^2]}{\Delta x} = \lim (2x + \Delta x)= 2x$.

But if we chose to say that the derivative is a value where $\Delta x$ is a number "very near zero" (I'm going to ignore the difficulty of how one would define such an ambiguous expression) so that $\Delta x = \delta >0$ but $\delta$ is "small".

Then $\dfrac {\Delta y}{\Delta x} = \dfrac {(x+\delta)^2 - x^2}{\delta} \dfrac {x^2 + 2x\delta + \delta^2 }{\delta} = 2x + \delta \ne 2x$.

So, no, that is a different answer. You can't say $\Delta x$ isn't zero at the beginning and then say "well $\Delta x$ very close to $0$ so we can ignore it" in the end.

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You can consider the derivative to be the shadow of the ratio $\frac{\Delta y}{\Delta x}$ when $\Delta x$ is infinitesimal. Alternatively, you can exploit the kind of relation Leibniz had in mind, which was a relation of infinite proximity. Leibniz wrote in his articles that when he speaks of equality, he does not mean strict equality but rather equality up to a negligible term.

Leibniz did not distinguish in notation between strict equality and equality up to a neglibigle infinitesimal, but he did use an alternative notation ${}_{\ulcorner\!\urcorner}\,$ for such a relation so it may be instructive to express this thought as follows. If $y=f(x)$ then one can write that $f'(x)\;{}_{\ulcorner\!\urcorner}\,\frac{\Delta y}{\Delta x}$ while understanding that $f'(x)$ has to be real (or, as Leibniz would have put it, "assignable").

Shadow is the same as the standard part.

Incidentally, Leibniz's remarks about a notion of equality "up to a negligible term" show that Bishop Berkeley's criticism of the calculus had no merit; see this article for the details.

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    $\begingroup$ I think deleting and downvoting this post is unreasonable, and am firmly against it. Yes it involves non-standard analysis, no that doesn't make it worthless. $\endgroup$ – Stella Biderman Apr 24 '17 at 16:39
  • $\begingroup$ @StellaBiderman I agree. I'm not at all claiming this user is correct, but he is definitely referring to historically sound concepts. If anything, the ideas are just a bit archaic for the modern interpretation of calculus. $\endgroup$ – The Great Duck Apr 24 '17 at 19:48
  • $\begingroup$ @TheGreatDuck, the fantasy that one can attach a definite article to the expression "modern interpretation of calculus" is precisely the misconception that we have addressed in a number of articles in leading venues; see here. $\endgroup$ – Mikhail Katz Apr 25 '17 at 7:23
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    $\begingroup$ Actually your views on Newton are archaic and need updating. Newton's notion of ultimate ratio was equivalent to limits. You can perform the following mental experiment: replace "lim" by "ult" in your favorite calculus definition, and let me know if that changes anything, e.g. $f'(x)=\text{ult}_{h\to 0}\frac{f(x+h)-f(x)}{h}$. Newton knew that ultimate ratios were not literally ratios. This is known to informed historians of mathematics. @TheGreatDuck $\endgroup$ – Mikhail Katz Apr 26 '17 at 7:02
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    $\begingroup$ @TheGreatDuck, the most popular definition of the tangent line to a curve in those days was "the line through two infinitely close points" on the curve. I never heard about your theory concerning the mean value theorem and I find it dubious. Anyway the main point here is that Newton already knew about limits under the name "ultimate ratios" which as he wrote were not true ratios. This was already noted by authorities like Bertrand Russell, and was emphasized recently by Bruce Pourciau. $\endgroup$ – Mikhail Katz Apr 27 '17 at 9:17

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