If y = (1 + frac{1}{x}) (1 + frac{2}{x}) (1 + | vedclass

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(C) Given $y = \prod_{k=1}^{n} (1 + \frac{k}{x}) = \prod_{k=1}^{n} (\frac{x+k}{x}) = \frac{(x+1)(x+2)...(x+n)}{x^n}$.Taking the natural logarithm on b

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$(-1)^n (n - 1)!$

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(C) Given $y = \prod_{k=1}^{n} (1 + \frac{k}{x}) = \prod_{k=1}^{n} (\frac{x+k}{x}) = \frac{(x+1)(x+2)...(x+n)}{x^n}$.Taking the natural logarithm on both sides:$\ln y = \sum_{k=1}^{n} \ln(1 + \frac{k}{x})$.Differentiating with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = \sum_{k=1}^{n} \frac{1}{1 + \frac{k}{x}} \cdot (-\frac{k}{x^2}) = \sum_{k=1}^{n} \frac{x}{x+k} \cdot (-\frac{k}{x^2}) = -\sum_{k=1}^{n} \frac{k}{x(x+k)}$.Thus,$\frac{dy}{dx} = -y \sum_{k=1}^{n} \frac{k}{x(x+k)}$.At $x = -1$,$y = (1-1)(1-2)...(1-n) = 0$ if $n \geq 1$. However,we must evaluate the limit or the product form carefully.Using the product rule: $\frac{dy}{dx} = y \sum_{k=1}^{n} \frac{d}{dx} \ln(1 + \frac{k}{x}) = y \sum_{k=1}^{n} \frac{-k}{x(x+k)}$.For $x = -1$,$y = (1-1)(1-2)...(1-n) = 0$. The term corresponding to $k=1$ in the sum is $\frac{-1}{x(x+1)} = \frac{-1}{-1(0)}$,which is undefined. Re-evaluating: $y = \frac{(x+1)(x+2)...(x+n)}{x^n}$.Using the product rule $\frac{d}{dx} [u_1 u_2 ... u_n] = \sum_{i=1}^{n} u_i' \prod_{j 
eq i} u_j$.At $x = -1$,only the term where $u_1 = (1 + \frac{1}{x}) = \frac{x+1}{x}$ is differentiated survives,because all other terms contain $(x+1)$.$\frac{dy}{dx} |_{x=-1} = (\frac{d}{dx} (1 + \frac{1}{x}))_{x=-1} \cdot (1 + \frac{2}{x}) (1 + \frac{3}{x}) ... (1 + \frac{n}{x}) |_{x=-1}$.$= (-\frac{1}{x^2})_{x=-1} \cdot (1-2)(1-3)...(1-n) = -1 \cdot (-1)^{n-1} (n-1)! = (-1)^n (n-1)!$.

If $y = (1 + \frac{1}{x}) (1 + \frac{2}{x}) (1 + \frac{3}{x}) . . . . . . (1 + \frac{n}{x})$ and $x \neq 0$. When $x = -1$,find $\frac{dy}{dx}$.

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