This post is about binomial distribution with standard normal distribution.
$$h(x):=\ln(x)~~\text{where}~~\underbrace{1\ll x\in\mathbb{N}}_{\text{very large natural number}}~~\text{is held}\tag{1}$$
$$\frac{\mathrm{d}}{\mathrm{dx}}\left(h(x)\right)\approx\frac{h(x)-h(x-\Delta x)}{\Delta x}\tag{2}$$
$$=\underbrace{\frac{h(x)-h(x-1)}{1}}_{\because~~\left(\Delta x=1\right)}\tag{3}$$
$$=\ln\left(x!\right)-\ln\left(\left(x-1\right)!\right)\tag{4}$$
$$=\ln\left(\frac{x!}{\left(x-1\right)!}\right)\tag{5}$$
$$=\ln\left(x\right)\tag{6}$$
$$n:=\text{very large natural number}\tag{7}$$
$$\mu_{}:=np~~\text{where}~~p~~\text{represents the probability of occur of the event}\tag{8}$$
$$x~~\text{exists around}~~\mu_{}=np\tag{9}$$
What I can't get currently is the following.
$$\color{red}{\frac{\mathrm{d}}{\mathrm{dx}}\left(\ln\left(\left(n-x\right)!\right)\right)\approx\ln\left(n-x\right)\cdot\frac{\mathrm{d}}{\mathrm{dx}}\left(\left(n-x\right)\right)}\tag{10}$$
Can anyone give me some hints?
What I want to say is that following should be held I think.
$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\ln\left(\left(n-x\right)!\right)\right)\approx\ln\left(\left(n-x\right)\right)\tag{11}$$
However if there is no misprint in the book, the following is held.
$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\ln\left(\left(n-x\right)!\right)\right)\approx-\ln\left(\left(n-x\right)\right)\tag{12}$$
Using a chain rule is enough.
$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\ln\left(x!\right)\right)\approx\ln\left(x\right)\tag{1}$$
$$u:=n-x\tag{2}$$
$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\ln\left(\left(n-x\right)!\right)\right)\tag{3}$$
$$=\frac{\mathrm{d}}{\mathrm{dx}}\ln\left(u!\right)\tag{4}$$
$$=\frac{\mathrm{d}}{\mathrm{du}}\ln\left(u!\right)\frac{\mathrm{du}}{\mathrm{dx}}\tag{5}$$
$$\approx\ln\left(u\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(n-x\right)\tag{6}$$
$$=-\ln\left(n-x\right)\tag{7}$$