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

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

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

Now define the statement

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

Base case: when \(n=3\),

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

so \(H_3\) is true.

Inductive step: assume \(H_n\) is true for some \(n\geqslant 3\), so

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

Differentiate both sides with respect to \(x\):

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

Simplifying,

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

This is exactly the required formula with \(n\) replaced by \(n+1\):

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

Hence, by mathematical induction, the formula is true for every integer \(n\geqslant 3\).