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

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Aicharya · Accepted Answer

$\begin{array}{ll} y= & \left((x+1)(4 x+1)(9 x+1) \ldots\left(\mathrm{n}^2 x+1ight)ight)^2 \ 	herefore \quad & \log y=2[\log (x+1)+\log (4 x+1)+\log (9 x+1) \ & \left.+\ldots+\log \left(\mathrm{n}^2 x+1ight)ight] \end{array}$ Differentiating w.r.t. $x$, we get $\frac{1}{y} \frac{\mathrm{~d} y}{\mathrm{~d} x}=2\left[\frac{1}{(x+1)}+\frac{4}{4 x+1}+\frac{9}{9 x+1}+\ldots+\frac{\mathrm{n}^2}{\mathrm{n}^2 \dot{x}+1}ight]$ $	herefore \quad \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}ight]$ $\begin{aligned}\left.	herefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}ight|_{x=0} & =2 y(0)\left[1^2+2^2+3^2+\ldots+\mathrm{n}^2ight] \ & =2(1) \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6} \ & =\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{3}\end{aligned}$

If $y=\left((x+1)(4 x+1)(9 x+1) \ldots\left(\mathrm{n}^2 x+1\right)\right)^2$, then…

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