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

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Aicharya · Accepted Answer

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

If $y=[(x+1)(2 x+1)(3 x+1) \ldots \ldots \ldots \ldots(\mathrm{n} x+1)]^4$ then $\frac{\mathrm{d}…

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