If
\(y=[(x+1)(2 x+1)(3 x+1) \ldots \ldots \ldots(\mathrm{n} x+1)]^2,\)
then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) at \(x=0\) is

\(y=[(x+1)(2 x+1)(3 x+1) \ldots(\mathrm{n} x+1)]^2\)
Taking 'log' on both sides, we get
\(\log y=2[\log (x+1)+\log (2 x +1)+\log (3 x+1)\) \(+\ldots+\log (n x+1)]\)
Differentiating w.r.t. \(\dot{x}\), we get
\(\frac{1}{y} \cdot \frac{\mathrm{~d} y}{\mathrm{~d} x}=2\left(\frac{1}{x+1}+\frac{2}{2 x+1}+\frac{3}{3 x+1}+\ldots+\frac{\mathrm{n}}{\mathrm{n} x+1}\right) \)
\(\therefore \frac{\mathrm{d} y}{\mathrm{~d} x}=2 y\left(\frac{1}{x+1}+\frac{2}{2 x+1}+\frac{3}{3 x+1}+\ldots+\frac{\mathrm{n}}{\mathrm{n} x+1}\right)\)
Now at \(x=0, y=[(1)(1)(1) \ldots(1)]^2=1\)
\(\therefore \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0} =2(1)\left(\frac{1}{0+1}+\frac{2}{0+1}+\frac{3}{0+1}+\ldots+\frac{\mathrm{n}}{0+1}\right) \)
\( =2(1+2+3+\ldots+\mathrm{n}) \)
\( =2 \times \frac{\mathrm{n}(\mathrm{n}+1)}{2}=\mathrm{n}(\mathrm{n}+1)\)