If \(y=[(x+1)(2 x+1)(3 x+1) \ldots \ldots \ldots \ldots(\mathrm{n} x+1)]^4\) then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) at \(x=0\) is

\(y=[(x+1)(2 x+1)(3 x+1) \ldots(n x+1)]^4 \)
\( \Rightarrow \log y=4[\log (x+1)(2 x+1)(3 x+1) \ldots(n x+1)] \)
\( \Rightarrow \log y=4[\log (x+1)+\log (2 x+1) \) \(+\log (3 x+1)+\ldots+\log (n x+1)]\)
Differentiating both sides w.r.t. \(x\), we get
\(\frac{1}{y} \frac{\mathrm{~d} y}{\mathrm{~d} x}=4\left[\frac{1}{x+1}+\frac{2}{2 x+1}+\frac{3}{3 x+1}+\ldots+\frac{\mathrm{n}}{\mathrm{n} x+1}\right] \)
\( \Rightarrow \frac{1}{1}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)_{x=0}=4(1+2+3+\ldots \mathrm{n}) \)
\( \Rightarrow\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0}=4\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)=2 \mathrm{n}(\mathrm{n}+1)\)