Answer: (i) \(\frac{\mathrm{d}^{n+1}}{\mathrm{d}x^{n+1}}(x^{n+1}\ln x)=\frac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}(x^{n}+(n+1)x^{n}\ln x)\).

(ii) For all positive integers \(n\), \(\frac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}(x^{n}\ln x)=n!\left(\ln x+1+\frac12+\cdots+\frac1n\right)\).

(i) Use the product rule on \(x^{n+1}\ln x\):

\(\frac{\mathrm{d}}{\mathrm{d}x}(x^{n+1}\ln x)=x^{n+1}\cdot\frac1x+(n+1)x^n\ln x=x^n+(n+1)x^n\ln x.\)

Now apply \(\frac{\mathrm{d}^n}{\mathrm{d}x^n}\) to both sides:

\(\frac{\mathrm{d}^{n+1}}{\mathrm{d}x^{n+1}}(x^{n+1}\ln x)=\frac{\mathrm{d}^n}{\mathrm{d}x^n}\bigl(x^n+(n+1)x^n\ln x\bigr).\)

This is the required result.

(ii) We prove by induction on \(n\) that

\(H_n:\quad \frac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}(x^{n}\ln x)=n!\left(\ln x+1+\frac12+\cdots+\frac1n\right).\)

Base case. For \(n=1\),

\(\frac{\mathrm{d}}{\mathrm{d}x}(x\ln x)=\ln x+1,\)

which agrees with \(1!\left(\ln x+1\right)\). So \(H_1\) is true.

Inductive step. Assume \(H_k\) is true for some positive integer \(k\), so

\(\frac{\mathrm{d}^{k}}{\mathrm{d}x^{k}}(x^{k}\ln x)=k!\left(\ln x+1+\frac12+\cdots+\frac1k\right).\)

Using part (i) with \(n=k\),

\(\frac{\mathrm{d}^{k+1}}{\mathrm{d}x^{k+1}}(x^{k+1}\ln x)=\frac{\mathrm{d}^{k}}{\mathrm{d}x^{k}}\bigl(x^{k}+(k+1)x^{k}\ln x\bigr).\)

Differentiate term by term:

\(\frac{\mathrm{d}^{k}}{\mathrm{d}x^{k}}(x^k)=k!,\)

and, by the inductive hypothesis,

\(\frac{\mathrm{d}^{k}}{\mathrm{d}x^{k}}(x^{k}\ln x)=k!\left(\ln x+1+\frac12+\cdots+\frac1k\right).\)

Hence

\(\frac{\mathrm{d}^{k+1}}{\mathrm{d}x^{k+1}}(x^{k+1}\ln x)=k!+(k+1)k!\left(\ln x+1+\frac12+\cdots+\frac1k\right).\)

Factor out \((k+1)!\):

\(= (k+1)!\left(\ln x+1+\frac12+\cdots+\frac1k+\frac1{k+1}\right).\)

This is exactly \(H_{k+1}\).

Therefore, since \(H_1\) is true and \(H_k\Rightarrow H_{k+1}\), the result holds for all positive integers \(n\).