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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 notice 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?

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 notice 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?

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