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Consider a function $F:\mathbb{R}\rightarrow \mathbb{R}$. I know this definition of derivative of $F$ at $x$:

$$ \lim_{h \rightarrow 0} \frac{F(x+h)-F(x)}{h} $$

I found this definition of derivative in a textbook

$$ \lim_{h \rightarrow 0} \frac{F(x+\frac{h}{2})-F(x-\frac{h}{2})}{h} $$

Could you help me to understand why they are equivalent?

Star
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2 Answers2

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They are not equivalent. With the second definition, $F(x) = |x|$ would be differentiable at $x=0$.

mrf
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Set $h^\prime \stackrel{\rm def}{=} \frac{h}{2}$. Then $$\begin{align*} \frac{F(x+\frac{h}{2})-F(x-\frac{h}{2})}{h}&= \frac{F(x+h^\prime) - F(x)+F(x)-F(x-h^\prime)}{2h^\prime} \\ &= \frac{1}{2}\left( \frac{F(x+h^\prime) - F(x)}{h^\prime}+\frac{F(x-h^\prime)-F(x)}{-h^\prime} \right) \end{align*}$$ Does that help seeing the relation with the first definition?

Note: the catch is that the second definition does not allow a function to be "differentiable from the left (resp. the right) only." You need the derivative on both sides of $x$ to exist and be equal for it to be equal to the first definition.

Clement C.
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  • So it's the average of the left and right derivatives (if those exist)? – Akiva Weinberger Jan 22 '16 at 13:50
  • You can see it like this, yes. In particular, if $F$ is differentiable at $x$, then the second expression will be equal to the first. – Clement C. Jan 22 '16 at 13:50
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    I think that for $f(x)=x\sin(\frac1x)$, the second expression exists despite the fact that it has neither a left nor a right derivative. Though I'll have to double-check. – Akiva Weinberger Jan 22 '16 at 13:55
  • Yes -- I edited a bit. What I wanted to say in the "note" is that the first expression can be modified to only take the limit when $h > 0$ (limit from the right) or $h < 0$ (limit from the left). You cannot have such distinction with the second. – Clement C. Jan 22 '16 at 13:58
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    @AkivaWeinberger You are correct. With the second "definition", every even function is differentiable at $x=0$. – mrf Jan 22 '16 at 13:58
  • @ClementC. I don't quite get what you mean in the note. Could you explain better? Thanks a lot – Star Jan 22 '16 at 15:44
  • By modifying very slightly the first definition, one can define the right-derivative of $F$ at $x$ as the limit (if it exists) of $\frac{F(x+h)-F(x)}{h}$ when $h\to 0^+$ (i.e., $h > 0$). Similarly, by choosing $h \to 0^-$ ($h<0$), one can define the left-derivative. But the second "definition" does not allow such refinements: for instance, if $F$ has a left-derivative equal to $0$ at $x$, but no right-derivative ($F(x)$ goes, say, to $\infty$ when $x\to x^+$), then there is no simple way to capture this with (a modification of) the second expression. – Clement C. Jan 22 '16 at 15:47
  • In a sense, the second "definition" is more general (see examples given in other answers and comments), but does not allow straightforwardly nice generalizations as the one above. – Clement C. Jan 22 '16 at 15:49
  • Still unclear: according to the first definition, $F(x)$ is differentiable at $x$ if $\exists \lim_{h\rightarrow 0}\frac{F(x+h)-F(x)}{h} \leftrightarrow \exists \lim_{h\rightarrow 0^+}\frac{F(x+h)-F(x)}{h}=\lim_{h\rightarrow 0^-}\frac{F(x+h)-F(x)}{h}$. According to the second definition, $F(x)$ is differentiable at $x$ if $\exists \lim_{h'\rightarrow 0}\frac{F(x+h')-F(x)}{h'} \leftrightarrow \exists \lim_{h'\rightarrow 0^+}\frac{F(x+h')-F(x)}{h'}=\lim_{h'\rightarrow 0^-}\frac{F(x+h')-F(x)}{h'}$ and .... – Star Jan 22 '16 at 16:03
  • $\exists \lim_{h'\rightarrow 0}\frac{F(x-h')-F(x)}{-h'} \leftrightarrow \exists \lim_{h'\rightarrow 0^+}\frac{F(x-h')-F(x)}{-h'}=\lim_{h'\rightarrow 0^-}\frac{F(x-h')-F(x)}{-h'}$. By noticing that $\lim_{h'\rightarrow 0^+}\frac{F(x-h')-F(x)}{-h'}=\lim_{h'\rightarrow 0^-}\frac{F(x+h')-F(x)}{h'}$ and $\lim_{h'\rightarrow 0^-}\frac{F(x-h')-F(x)}{-h'}=\lim_{h'\rightarrow 0^+}\frac{F(x+h')-F(x)}{h'}$,... – Star Jan 22 '16 at 16:04
  • we conclude that, according to the second definition, $F(x)$ is differentiable at $x$ if $\exists \lim_{h'\rightarrow 0}\frac{F(x+h')-F(x)}{h'} \leftrightarrow \exists \lim_{h'\rightarrow 0^+}\frac{F(x+h')-F(x)}{h'}= \lim_{h'\rightarrow 0^-}\frac{F(x+h')-F(x)}{h'}$, which is equivalent to the requirement of the first definition. – Star Jan 22 '16 at 16:05