[(-1) (n-1)] [(n-1) factorial ]/( 1+x) n fixed formula

Just expand it with Leibniz. In the end, only two items are not zero, and the others are all zero.

One term is the n-order derivative of ln (1+x)

The other term is the n-3 derivative of c (n 3) * (x 3)' * ln (1+x).

C (n 3) stands for permutation and combination, which means that taking 3 from n' stands for the first derivative.

Just add these two items together and bring in x=0.

There are several other items, and the last one with an X gets zero, so I didn't write down the 4 5 6 of X. There are always four terms in the formula with all zero derivatives, two of which I didn't list with X.

I can solve this problem with Taylor's formula. If I can't write well, I won't write.

What is said upstairs is right. You just need to take the derivative of x at the same time. Obviously, there will be a result. Do the math yourself.