If y = ((x+1)(4x+1)(9x+1) ldots (n^2x+1))^2,t | vedclass

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(D) Given $y = ((x+1)(4x+1)(9x+1) \ldots (n^2x+1))^2$.Taking the natural logarithm on both sides:$\log y = 2[\log(x+1) + \log(4x+1) + \log(9x+1) + \ld

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$\frac{n(n+1)(2n+1)}{3}$

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(D) Given $y = ((x+1)(4x+1)(9x+1) \ldots (n^2x+1))^2$.Taking the natural logarithm on both sides:$\log y = 2[\log(x+1) + \log(4x+1) + \log(9x+1) + \ldots + \log(n^2x+1)]$.Differentiating with respect to $x$ using the chain rule:$\frac{1}{y} \frac{dy}{dx} = 2 \left[ \frac{1}{x+1} + \frac{4}{4x+1} + \frac{9}{9x+1} + \ldots + \frac{n^2}{n^2x+1} ight]$.Thus,$\frac{dy}{dx} = 2y \left[ \frac{1^2}{x+1} + \frac{2^2}{4x+1} + \frac{3^2}{9x+1} + \ldots + \frac{n^2}{n^2x+1} ight]$.At $x=0$,$y(0) = ((0+1)(0+1) \ldots (0+1))^2 = 1^2 = 1$.Substituting $x=0$ into the derivative expression:$\left. \frac{dy}{dx} ight|_{x=0} = 2(1) [1^2 + 2^2 + 3^2 + \ldots + n^2]$.Using the sum of squares formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$:$\left. \frac{dy}{dx} ight|_{x=0} = 2 	imes \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(2n+1)}{3}$.

If $y = ((x+1)(4x+1)(9x+1) \ldots (n^2x+1))^2$,then $\frac{dy}{dx}$ at $x=0$ is

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