If \(y=\left((x+1)(4 x+1)(9 x+1) \ldots\left(\mathrm{n}^2 x+1\right)\right)^2\), then \(\frac{\mathrm{dy}}{\mathrm{d} x}\) at \(x=0\) is

\(y= \left((x+1)(4 x+1)(9 x+1) \ldots\left(\mathrm{n}^2 x+1\right)\right)^2 \)
\( \therefore \log y=2[\log (x+1)+\log (4 x+1)+\log (9 x+1) \) \( \left.+\ldots+\log \left(\mathrm{n}^2 x+1\right)\right]\)
Differentiating w.r.t. \(x\), we get
\(\frac{1}{y} \frac{\mathrm{~d} y}{\mathrm{~d} x}=2\left[\frac{1}{(x+1)}+\frac{4}{4 x+1}+\frac{9}{9 x+1}+\ldots+\frac{\mathrm{n}^2}{\mathrm{n}^2 \dot{x}+1}\right]\)
\(\therefore \frac{\mathrm{d} y}{\mathrm{~d} x}=2 y\left[\frac{(1)^2}{x+1}+\frac{(2)^2}{4 x+1}+\frac{(3)^2}{9 x+1}+\ldots+\frac{\mathrm{n}^2}{\mathrm{n}^2 x+1}\right]\)
\(\left.\therefore \frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=0} =2 y(0)\left[1^2+2^2+3^2+\ldots+\mathrm{n}^2\right] \)
\( =2(1) \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6} \)
\( =\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{3}\)