If y = [(x+1)(2x+1)(3x+1) ldots (nx+1)]^2,the | vedclass

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(B) Given $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^2$. Taking the natural logarithm on both sides,we get: $\ln y = 2[\ln(x+1) + \ln(2x+1) + \ln(3x+1) +

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$n(n+1)$

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(B) Given $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^2$. Taking the natural logarithm on both sides,we get: $\ln y = 2[\ln(x+1) + \ln(2x+1) + \ln(3x+1) + \ldots + \ln(nx+1)]$. Differentiating both sides with respect to $x$,we get: $\frac{1}{y} \cdot \frac{dy}{dx} = 2 \left[ \frac{1}{x+1} + \frac{2}{2x+1} + \frac{3}{3x+1} + \ldots + \frac{n}{nx+1} ight]$. Therefore,$\frac{dy}{dx} = 2y \left[ \frac{1}{x+1} + \frac{2}{2x+1} + \frac{3}{3x+1} + \ldots + \frac{n}{nx+1} ight]$. At $x=0$,$y = [(0+1)(0+1)(0+1) \ldots (0+1)]^2 = 1^2 = 1$. Substituting $x=0$ and $y=1$ into the expression for $\frac{dy}{dx}$: $\left( \frac{dy}{dx} ight)_{x=0} = 2(1) \left[ \frac{1}{0+1} + \frac{2}{0+1} + \frac{3}{0+1} + \ldots + \frac{n}{0+1} ight]$. $= 2(1 + 2 + 3 + \ldots + n)$. Using the sum formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$,we get: $= 2 	imes \frac{n(n+1)}{2} = n(n+1)$.

If $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^2$,then $\frac{dy}{dx}$ at $x=0$ is

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