$$
\begin{aligned}
& y=[(x+1)(2 x+1)(3 x+1) \ldots(\mathrm{n} x+1)]^{\mathrm{n}} \\
& \Rightarrow \log y=\operatorname{nlog}[(x+1)(2 x+1)(3 x+1) \\
& \Rightarrow \log y=\mathrm{n}[\log (x+1)+\log (2 x+1) \\
& \quad+\log (3 x+1)+\ldots+\log (\mathrm{n} x+1)]
\end{aligned}
$$
Differentiating both sides w.r.t. $x$, we get
$$
\begin{aligned}
& \frac{1}{y} \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{n}\left(\frac{1}{x+1}+\frac{2}{2 x+1}+\frac{3}{3 x+1}+\ldots+\frac{\mathrm{n}}{\mathrm{n} x+1}\right) \\
& \Rightarrow \frac{1}{1} \cdot\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0}=\mathrm{n}(1+2+3+\ldots+\mathrm{n})
\end{aligned}
$$
$\ldots[$ At $x=0, y=1]$
$\Rightarrow\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0}=\mathrm{n}\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]=\frac{\mathrm{n}^2(\mathrm{n}+1)}{2}$