Using Leibniz to find the fourth derivative of x ^ 3 · LNX

Using Leibniz to find the fourth derivative of x ^ 3 · LNX

The fourth derivative of x ^ 3 = 3! = 6, the first derivative of LNX = 1 / XX ^ 3, the second derivative of LNX = 6x, the second derivative of LNX = - 1 / x ^ 2x ^ 3, the first derivative of LNX = 3x ^ 2, the third derivative of LNX = 2X ^ (- 3) x ^ 3, the fourth derivative of LNX = - 6x ^ (- 4), so the original formula = C (4,0) × 0 + C (4,1) × 6 × 1 / x + C (4,2)

Who can explain the Leibniz formula of n-order derivative? That is, find the n-order derivative of UV (UV) ^ (n) = Σ Cu Λ (n-k) × V Λ K, where the superscript of C is K and the subscript of C is n. I can't understand it completely in the book. Because it's a self-taught high number, I hope to explain it in detail

Here is mainly to examine the high-order derivatives. You only need to remember a few commonly used high-order derivatives,
And remember the Leibniz formula, which doesn't examine the whole, because there's a trick to just do a few steps
I play a question, but is the picture, with the mobile phone can see clearly? No picture is difficult to type out, and not intuitive

Sum: 1 + 2x + 3x ＋nx∧（n－1）

Sn=1+2x+3x^2… +nx^(n-1) ①
When x = 1, Sn = 1 + 2 + 3 + +n=（n+1）n/2
If ^ 2x + 2 ≠ x ≠ +nx^n ②
①-②:(1-x)Sn=1+x+x^2+x^(n-1)-nx^n
=(1-x^n)/(1-x) -nx^n
∴Sn=(1-x^n)/(1-x)^2 -nx^n/(1-x)