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I have a question which asks me if the following two definitions of a derivative is equal. So I know the following equation,

$f'(x_0) = \lim_{h \to 0} \dfrac{f(x_0+h) - f(x_0)}{h} $

as we went through it when I learnt this and how to get the derivative using limits as h goes to $0$ but I don't get the following equation.

$g'(x_0) = \lim_{h \to 0} \dfrac{g(x_0+h) - g(x_0-h)}{2h} $

The question is asking me knowing that we defined a derivative of $\ f$ at $\ x_0$, is the following definition suggested by someone as equivalent and if so why? How would I go about answering this? Super confused about where to start?

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Consider the function $g(x) = |x|$ and $x_0 =0$ or the function $g(x) = 1/x^2$ and $x_0 = 0$ (not in the domain). The second "definition" would imply $g'(0) = 0$.

If the second "definition" were true, then the value of the function at $x_0$ would be irrelevant - hence the "derivative" at $x_0$ might exist even if the function were not continuous, or not even defined at $x_0$.

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Remember the derivative is the slope of the tangent line. This is obtained by taking a limit of slopes of secant lines. So, if you consider the points $(x-h, f(x-h))$ and $(x+h, f(x+h))$, using the slope formula gives $$ m = \frac{f(x+h) - f(x-h)}{(x+h) - (x-h)} $$ which of course cleans up as $$ m = \frac{f(x+h) - f(x-h)}{2h}. $$ Now pass $h \to 0.$

In short, yes, the definitions are equivalent.

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    $\begingroup$ @Catalin Zara has given an example that shows they are NOT equivalent. The alternate formulation is known as the symmetric derivative, and quite a bit of information can be found by googing. For a brief historical survey of certain aspects of this notion (that one day I'll try to write up more completely as an answer), see the later comments of mine at If $\,\lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}\,$ exists for every $x$, what does this imply for $f$? $\endgroup$ – Dave L. Renfro Oct 15 '18 at 0:00
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    $\begingroup$ For a geometric interpretation of the symmetric derivative and similar notions, see my comments in the paragraph beginning with As to your question about “unstraddled difference quotient” approaches in my answer to “Strong” derivative of a monotone function. Also worth looking at, given what you wrote, are the places I cited in Fowler's book The Elementary Differential Geometry of Plane Curves. $\endgroup$ – Dave L. Renfro Oct 15 '18 at 0:05
  • $\begingroup$ @Dave L. Renfro But wouldn't the same rules apply for differentiability at $x_0$. That is, continuity, left derivative equals right derivative etc. The first definition doesn't take into account the left side derivative. Does adding the statement, "where the derivative exists" make them equivalent? $\endgroup$ – Phil H Oct 15 '18 at 1:19
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    $\begingroup$ @Phil H: Does adding the statement, "where the derivative exists" make them equivalent? --- If the ordinary derivative exists, then the symmetric derivative exists. However, the symmetric derivative can exist when the ordinary derivative doesn't exist. Thus, the symmetric derivative is a weaker notion of being differentiable than the usual notion. The whole point of my various comments I linked to has to do with how many points is it possible for a function that is everywhere symmetrically differentiable to not be ordinarily differentiable, where "how many" is measured in various ways. $\endgroup$ – Dave L. Renfro Oct 15 '18 at 1:33
  • $\begingroup$ In the light of comments made by @DaveL.Renfro you should update your answer. $\endgroup$ – Paramanand Singh Oct 15 '18 at 9:42

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