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

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(B) Let $f(x) = (x+1)(2x+1)(3x+1) \dots (nx+1)$. Then $y = [f(x)]^4$.Using the chain rule, $\frac{dy}{dx} = 4[f(x)]^3 \cdot f'(x)$.At $x = 0$, $f(0) =

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$n(n+1)$

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(B) Let $f(x) = (x+1)(2x+1)(3x+1) \dots (nx+1)$. Then $y = [f(x)]^4$.Using the chain rule, $\frac{dy}{dx} = 4[f(x)]^3 \cdot f'(x)$.At $x = 0$, $f(0) = (1)(1)(1) \dots (1) = 1$.To find $f'(0)$, we use logarithmic differentiation: $\ln f(x) = \sum_{i=1}^{n} \ln(ix+1)$.Differentiating both sides: $\frac{f'(x)}{f(x)} = \sum_{i=1}^{n} \frac{i}{ix+1}$.At $x = 0$, $\frac{f'(0)}{f(0)} = \sum_{i=1}^{n} i = \frac{n(n+1)}{2}$.Since $f(0) = 1$, $f'(0) = \frac{n(n+1)}{2}$.Now, $\frac{dy}{dx} \big|_{x=0} = 4[f(0)]^3 \cdot f'(0) = 4(1)^3 \cdot \frac{n(n+1)}{2} = 2n(n+1)$.Given $\frac{dy}{dx} \big|_{x=0} = 2k$, we have $2k = 2n(n+1)$, so $k = n(n+1)$.

If $y = [(x+1)(2x+1)(3x+1) \dots (nx+1)]^4$, where $n \in N$ and $\frac{dy}{dx}$ at $x = 0$ is $2k$, then the value of $k$ is

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