$y=[(x+1)(2 x+1)(3 x+1) \ldots(\mathrm{n} x+1)]^{\frac{3}{2}}$
Taking 'log' on both sides, we get
$\begin{array}{r}
\log y=\frac{3}{2}[\log (x+1)+\log (2 x+1)+\log (3 x+1) \\
+\ldots+\log (n x+1)]
\end{array}$
Differentiating w.r.t. $x$, we get
$\begin{aligned}
& \quad \frac{1}{y} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3}{2}\left[\frac{1}{x+1}+\frac{2}{2 x+1}+\frac{3}{3 x+1}+\ldots+\frac{\mathrm{n}}{\mathrm{n} x+1}\right] \\
& \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3 y}{2}\left[\frac{1}{x+1}+\frac{2}{2 x+1}+\frac{3}{3 x+1}+\ldots+\frac{\mathrm{n}}{\mathrm{n} x+1}\right] \\
& \text { Now at } x=0, y=[\underbrace{(1)(1)(1) \ldots(1)}_{\mathrm{n} \text { times }}]^{\frac{3}{2}}=1
\end{aligned}$
$\begin{aligned}\left.\therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=0} & =\frac{3(1)}{2}\left[\frac{1}{0+1}+\frac{2}{0+1}+\frac{3}{0+1}+\ldots+\frac{\mathrm{n}}{0+1}\right] \\ & =\frac{3}{2}(1+2+3+\ldots+\mathrm{n}) \\ & =\frac{3}{2} \times \frac{\mathrm{n}(\mathrm{n}+1)}{2} \\ & =\frac{3 \mathrm{n}(\mathrm{n}+1)}{4}\end{aligned}$