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

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

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

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