Answer: (a) \(\frac{\mathrm{d}^3y}{\mathrm{d}x^3}=\frac{2}{1+x}\).

(b) For every integer \(n\geqslant 3\), \(\frac{\mathrm{d}^ny}{\mathrm{d}x^n}=(-1)^{n-1}\frac{2(n-3)!}{(1+x)^{n-2}}\).

Let \(y=(1+x)^2\ln(1+x)\).

Differentiate once using the product rule:

\(y' = 2(1+x)\ln(1+x) + (1+x)^2\cdot \frac{1}{1+x} = 2(1+x)\ln(1+x) + (1+x)\).

Differentiate again:

\(y'' = 2\ln(1+x) + 2 + 1 = 2\ln(1+x) + 3\).

Differentiate a third time:

\(y''' = 2\cdot \frac{1}{1+x} = \frac{2}{1+x}\).

So \(\frac{\mathrm{d}^3y}{\mathrm{d}x^3}=\frac{2}{1+x}\).

Now define the statement

\(H_n: \frac{\mathrm{d}^ny}{\mathrm{d}x^n}=(-1)^{n-1}\frac{2(n-3)!}{(1+x)^{n-2}}\), for \(n\geqslant 3\).

First check the base case \(n=3\):

\(\frac{\mathrm{d}^3y}{\mathrm{d}x^3}=\frac{2}{1+x}=(-1)^2\frac{2\cdot 0!}{(1+x)^1}\), so \(H_3\) is true.

Assume that for some integer \(k\geqslant 3\),

\(\frac{\mathrm{d}^ky}{\mathrm{d}x^k}=(-1)^{k-1}\frac{2(k-3)!}{(1+x)^{k-2}}\).

Differentiate both sides:

\(\frac{\mathrm{d}^{k+1}y}{\mathrm{d}x^{k+1}} = (-1)^{k-1}2(k-3)!\cdot \frac{\mathrm{d}}{\mathrm{d}x}\big((1+x)^{-(k-2)}\big)\).

Since

\(\frac{\mathrm{d}}{\mathrm{d}x}\big((1+x)^{-(k-2)}\big)=-(k-2)(1+x)^{-(k-1)}\),

we get

\(\frac{\mathrm{d}^{k+1}y}{\mathrm{d}x^{k+1}} = (-1)^k\frac{2(k-2)(k-3)!}{(1+x)^{k-1}} = (-1)^k\frac{2(k-2)!}{(1+x)^{k-1}}\).

This is exactly the required formula with \(n=k+1\):

\(\frac{\mathrm{d}^{k+1}y}{\mathrm{d}x^{k+1}} = (-1)^{(k+1)-1}\frac{2((k+1)-3)!}{(1+x)^{(k+1)-2}}\).

Therefore, by mathematical induction,

\(\frac{\mathrm{d}^ny}{\mathrm{d}x^n}=(-1)^{n-1}\frac{2(n-3)!}{(1+x)^{n-2}}\)

for every integer \(n\geqslant 3\).