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

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Aicharya · Accepted Answer

$y=[(x+1)(2 x+1)(3 x+1) \ldots(\mathrm{n} x+1)]^2$
Taking 'log' on both sides, we get $\begin{aligned} \log y=2[\log (x+1)+\log (2 x & +1)+\log (3 x+1) \\ & +\ldots+\log (n x+1)] \end{aligned}$ Differentiating w.r.t. $\dot{x}$, we get $\begin{aligned} & \frac{1}{y} \cdot \frac{\mathrm{~d} y}{\mathrm{~d} x}=2\left(\frac{1}{x+1}+\frac{2}{2 x+1}+\frac{3}{3 x+1}+\ldots+\frac{\mathrm{n}}{\mathrm{n} x+1}\right) \\ \therefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=2 y\left(\frac{1}{x+1}+\frac{2}{2 x+1}+\frac{3}{3 x+1}+\ldots+\frac{\mathrm{n}}{\mathrm{n} x+1}\right) \end{aligned}$
Now at $x=0, y=[(1)(1)(1) \ldots(1)]^2=1$ $\begin{aligned} \therefore \quad\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0} & =2(1)\left(\frac{1}{0+1}+\frac{2}{0+1}+\frac{3}{0+1}+\ldots+\frac{\mathrm{n}}{0+1}\right) \\ & =2(1+2+3+\ldots+\mathrm{n}) \\ & =2 \times \frac{\mathrm{n}(\mathrm{n}+1)}{2}=\mathrm{n}(\mathrm{n}+1) \end{aligned}$

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

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