<s>[INST] How to prove that problem $\frac{\partial ^m f_n(x)}{\partial x^m}$? Let $n,N\in\mathbb{N}$ and $x\in\mathbb{R}$.
Let $f_{n}(x)=(1-\frac{x}{n})^n-(1-\frac{x}{N})^{N}$.
Prove the following:
If ${N}≤n$ then
$$\frac{\partial ^{M}f_{n}}{\partial  x^M}(x)=(-1)^{M}\displaystyle\Pi_{k=0}^{M-1}\left(1-\frac{k}{n}\right)\left(1-\frac{x}{n}\right)^{n-M} \\+(-1)^{M-1}\Pi_{k=0}^{M-1}\left(1-\frac{k}{N}\right)\left(1-\frac{x}{N}\right)^{N-M}.$$
I tried to calculate $\frac{\partial^m f_n(x)}{\partial  x^m}$ where $f_n(x)=(1-\frac{x}{n})^n-(1-\frac{x}{b})^b$ if $b≤n$. I found
$$\frac{\partial f_{n}}{\partial x}(x)=-(1-\frac{x}{n})^{n-1}+(1-\frac{x}{b})^{b-1},$$
$$\frac{\partial^{2} f_{n}}{\partial x^{2}}=(\frac{n-1}{n})(1-\frac{x}{n})^{n-2}-(\frac{b-1}{b})(1-\frac{x}{b})^{b-2},$$
$$\frac{\partial^{3}f_{n}}{\partial x^3}(x)=(-1)^{3} (1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{x}{n})^{n-3} +\\
(-1)^{2=3-1} (1-\frac{1}{b})(1-\frac{2}{b})(1-\frac{x}{b})^{b-3}.$$
 [/INST] We put $f_n(x)=\sigma_1(x)-\sigma_2(x)$  where $\sigma_1(x)=(1-\frac{x}{n})^n$ and $ \sigma_2(x)=(1-\frac{x}{N})^N$
then we calculate $\frac{\partial ^M\sigma_1(x)}{\partial x^M}, \frac{\partial^M\sigma_2(x)}{\partial x^M }$
$\frac{\partial \sigma_1(x)}{\partial x}=(-1)^1(1-\frac{x}{n})^{n-1}$
$\frac{\partial^2\sigma_1(x)}{\partial x^2}=(-1)^2\frac{n-1}{n}(1-\frac{x}{n})^{n-2}$
$\frac{\partial^3\sigma_1(x)}{\partial x^3}=(-1)^3(\frac{1}{n})^2(n-1)(n-2)(1-\frac{x}{n})^{n-3}$
$\frac{\partial ^4\sigma_1(x)}{\partial x^4} = (-1)^4(\frac{1}{n^3}(n-1)(n-2)(n-3)(1-\frac{x}{n})^{n-4}$
So, we can now easily deduce the $m$-order derivative, starting from the first derivatives and we find:
$\frac{\partial ^M \sigma_1(x)}{\partial x^M }=(-1)^{M}(\frac{1}{n})^{M-1}(n-1)(n-2).......(n-(M-1))(1-\frac{x}{n})^{n-M}=(-1)^M\prod_{k=0}^{k=M-1}(1-\frac{k}{n})(1-\frac{x}{n})^{n-M}$
In the same way, we find:
$\frac{\partial ^M \sigma_2(x)}{\partial x^M}=(-1)^M\prod _{k=0}^{k=M-1}(1-\frac{k}{N})(1-\frac{x}{N})^{n-N}$
So :
$\frac{\partial ^M f_n(x)}{\partial x^M }=(-1)^M\prod_{k=0}^{k=M-1}(1-\frac{k}{n})\left((1-\frac{x}{n})^{n-M}-(1-\frac{x}{N})^{n-N}\right)$ .
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