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

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(B) Given $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^n$. Taking the natural logarithm on both sides: $\log y = n [\log(x+1) + \log(2x+1) + \log(3x+1) + \l

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$\frac{n^2(n+1)}{2}$

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

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

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