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I've been thinking about it since yesterday and have noticed this pattern:

We have, the first order derivative of a function $f(x)$ is: $$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} .......(1)$$
The second order derivative of the same function is: $$f''(x)=\lim_{h\rightarrow 0}\frac{f'(x+h)-f'(x)}{h}$$ By putting $x=x+h$ in (1), we can have $f'(x+h)$. So,$$f''(x)=\lim_{h\rightarrow 0}\frac{\lim_{h\rightarrow 0}\frac{f(x+h+h)-f(x+h)}{h}-\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}}{h}$$ Or , $$f''(x)=\lim_{h\rightarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}.....(2)$$ You can check by L'Hospital's rule that this limit evaluates to $f''(x)$. Now, the third order derivative of the same function is: $$f'''(x)=\lim_{h\rightarrow 0}\frac{f''(x+h)-f''(x)}{h}$$ By putting $x=x+h$ in (2), we can get $f''(x+h)$.

So, $$f'''(x)=\lim_{h\rightarrow 0}\frac{\lim_{h\rightarrow 0}\frac{f(x+3h)-2f(x+2h)+f(x+h)}{h^2}-\lim_{h\rightarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}}{h}$$ which gives $$f'''(x)=\lim_{h\rightarrow 0} \frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^3}......(3)$$ Again, by repeating the same process, we can get that: $$f''''(x)=\lim_{h\rightarrow 0}\frac{f(x+4h)-4f(x+3h)+6f(x+2h)-4f(x+h)+f(x)}{h^4}....(4)$$ So, we observe that the coefficient of $f(x+(n-r)h)$ in the expression of $f^{n}(x)$ ($n^{th}$ derivative of $f(x)$) is actually $(-1)^{r}\cdot {^n}C_r$, same as the coefficient of $x^{n-r}$ in the expansion of $(x-1)^n$. It can be proved that: $$f^n(x)=\lim_{h\rightarrow 0}\frac{\sum_{r=0}^n(-1)^{r}\cdot ^{n}C_r\cdot f(x+(n-r)h)}{h^n}$$ where $f^n(x)$ is the $n^{th}$ order derivative of the function $f(x)$.

Now, to generalize this to fractional order derivatives, we just have to generalize the coefficients, which must be similar to the generalization of the expansion of $(x-1)^n$ to fractional exponents.

I'm not very good with binomial theorem, but I guess that it should be: $$f^n(x)=\lim_{h\rightarrow 0}\frac{f(x+nh)-n\cdot f(x+(n-1)h)+\frac{n(n-1)}{2!}\cdot f(x+(n-2)h)-....}{h^n}$$ , where $n$ can be fractional. Have I done anything wrong?

UPDATE: A lot of people are saying that my method to get the expressions of $f''(x)$, $f'''(x)$, $f''''(x)$, etc are wrong. I've checked by L'Hospital's rule that the results are correct. Could you please write an answer about getting to these results by using a correct method?

UPDATE: On second thoughts, the $n^{th}$ derivative of a function can also proved to be equal to: $$\lim_{h\rightarrow 0^+}\frac{\sum_{r=0}^{n}(-1)^r\cdot \binom{n}{r}\cdot f(x-(n-r)h)}{(-h)^n}$$ If we try to expand this to fractions, it can turn out to be imaginary because of the fractional power of $-1$ in the denominator.

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    Very intriguing. Doesn't also the sum be adj for fractional $n$? And what about the $n$ times $h$ terms inside of $f$, i.e. $nh, (n-1)h,\ldots$, these would also need to be extended to fractions? – 2017-02-28
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    In your derivation of (2), how can you assume that $$f''(x) = \lim_{h\to 0} \frac{\lim_{h_1\to 0} \frac{f(x + h_1) - f(x)}{h_1} -\lim_{h_2 \to 0} \frac{f(x+h_2) - f(x)}{h_2}}{h}$$ simplifies to what you wrote? i.e: how do you justify taking all the convergence of $h\to 0$ for each individual limit? – 2017-02-28
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    @ZainPatel I don't understand why you're using two different $h_1$ and $h_2$. BTW, I made a mistake. I've edited the formula. – 2017-02-28
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    @ZainPatel My formula for $f''(x)$ is correct. You can check that by L'Hospital rule by twice differentiating both numerator and denominator. – 2017-02-28
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    $(2)$ is correct, but your derivation is flawed. – 2017-02-28
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    @Dove I know the result is correct, I'm just not convinced that your derivation works. – 2017-02-28
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    @ZainPatel You're correct. The result is correct; the development is not. – 2017-02-28
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    @Dove The $h$ variables in the limits in the numerator are *not* the same between them, and *not* the same as the $h$ in the limit of the fraction itself. You can't assume them to be the same and interchange limits like that, at least not without some solid argumentation. For a simplistic example, applying your "rule" to $(1)$ would give the obviously false: $$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} \;\;\ne\;\; \lim_{h\rightarrow 0}\frac{\lim_{h \to 0}\big(f(x+h)-f(x)\big)}{\lim_{h \to 0} h} = \frac{0}{0} \;\;\text{(undefined)}$$ – 2017-02-28
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    @Dr.MV I'm sorry. I can't see any flaw in it. – 2017-02-28
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    @ZainPatel I've never heard about using different $h's$ for different limits. – 2017-02-28
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    @Dove Taking the limits as $h_1\to 0$ and $h_2\to 0$ renders the numerator independent of $h$. – 2017-02-28
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    @Dove And you can't do that in *your* argument because (leaving the outer limit aside), the numerator evaluates to $f'(x)-f'(x)=0\,$. Even if defendable in *some* special cases, but you cannot hand-wave that in general without solid arguments, which is what I said before. – 2017-02-28
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    @Dr.MV So, what's a proper way to get to these results? – 2017-02-28
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    @Dove I've posted an answer to your question in the last comment. – 2017-02-28
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    If you generalize to fractional $n$ just like that, you end up with an infinite series which may face convergence problems. Besides, it takes infinitely many values of $f$ over an infinite range, so the derivative is no longer a _local_ property, which doesn't feel quite right. – 2017-02-28

2 Answers 2

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While the following does not address fractional derivatives, the OP has requested in a comment and in an UPDATE in the OP, a way to proceed rigorously to show that $$f''(x)=\lim_{h\to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}$$It is to that end only that we now proceed.

From the extended law of the mean, if $f$ is twice differentiable in the neighborhood of $x$, then for $h$ sufficiently small, there exists a number $\theta\in (0,1)$ such that

$$f(x+h)=f(x)+f'(x)h+\frac12 f''(x+\theta h)h^2 \tag 1$$

Alongside this, the extended law of the mean guarantees that there exists a number $\eta\in(0,1)$ such that

$$f(x-h)=f(x)-f'(x)h+\frac12 f''(x+\eta h)h^2 \tag 2$$

Adding $(1)$ and $(2)$ reveals

$$\frac12(f''(x+\theta h)+f''(x+\eta h))=\frac{f(x+h)-2f(x)+f(x-h)}{h^2} \tag 3$$

If $f''$ is continuous in a neighborhood of $x$, then we have from $(3)$

$$f''(x)=\lim_{h\to 0}\frac12(f''(x+\theta h)+f''(x+\eta h))=\lim_{h\to 0}\frac{f(x+h)-f(x)-f'(x)h}{h^2}$$

And we are done!

To obtain limit expressions for higher order derivatives, we simply use the extended mean value theorem

$$f(x+h)=f(x)+f'(x)h+\frac12f''(x)h^2+\cdots +f^{(n)}(x+\theta h)h^n \tag 4$$

and judiciously select different values combinations of $(4)$ to eliminate all derivative terms. This is left as an exercise for the reader.

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    Yes, they are the same. And the expressions for higher order derivatives can be validated from using the extended law of the mean. And L'Hospital's Rule is fine to validate their correctness too. So, well done! – 2017-02-28
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As an answer to the 'fractional derivative' part I mention that your derivation was on the way to Gruenwald-Leitnikov derivative. Notice the similarity in the approach and the summation :)

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    I guess they are the same. – 2017-03-01