If y = [(x+1)(2x+1)(3x+1) ldots (nx+1)]^4,the | vedclass

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(D) Given $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^4$.Taking logarithm on both sides:$\log y = 4[\log(x+1) + \log(2x+1) + \log(3x+1) + \ldots + \log(nx+

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$2n(n+1)$

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(D) Given $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^4$.Taking logarithm on both sides:$\log y = 4[\log(x+1) + \log(2x+1) + \log(3x+1) + \ldots + \log(nx+1)]$.Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = 4 \left[ \frac{1}{x+1} + \frac{2}{2x+1} + \frac{3}{3x+1} + \ldots + \frac{n}{nx+1} \right]$.At $x=0$,$y = [1 \cdot 1 \cdot 1 \ldots 1]^4 = 1$.Substituting $x=0$ into the derivative expression:$\frac{1}{1} \left( \frac{dy}{dx} \right)_{x=0} = 4 [1 + 2 + 3 + \ldots + n]$.Using the sum formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$:$\left( \frac{dy}{dx} \right)_{x=0} = 4 \left( \frac{n(n+1)}{2} \right) = 2n(n+1)$.

If $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^4$,then $\frac{dy}{dx}$ at $x=0$ is

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