Find the n-order derivative (n ≥ 3) of y = (x ^ 2) ln (1 + x) at x = 0

Find the n-order derivative (n ≥ 3) of y = (x ^ 2) ln (1 + x) at x = 0

It's very simple. I don't know that you have learned the knowledge of n-order derivative of Leibniz. If you don't know it very well, you can look up the relevant literature. I assume that you already know it
Given y = (x ^ 2) ln (1 + x)
The derivative of Y1 = x ^ 2, Y1 '= 2x, Y1' '= 2, Y1' '= 0
The n-order derivative of y2 = ln (1 + x) is Y2 ^ (n) = (- 1) ^ (n-1) * [(n-1)! / (1 + x) ^ n]
Because of Leibniz's derivative formula, here I specially stipulate that n is a positive integer greater than or equal to 1, y ^ (n) represents the derivative of order n, and others are similar
y^(n)=(y1*y2)^(n)=(x^2)[ln(1+x)]^(n)+n*(2x)[ln(1+x)]^(n-1)+[n(n-1)/2]*2*[ln(1+x)]^(n-2)
Sort it out
y^(n)=(-1)^(n-1)*(n-3)!*{[(3n+1-n^2)x^2+2(4n-3n^2)x+4n(n-1)]/(1+x)^(n)}
Where n is a positive integer greater than or equal to 1