We are tasked with differentiating the given function $y = \prod_{i=1}^N (x_i + 1)^2$ with respect to an arbitrary variable $x_m$. Here's the solution step-by-step:

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Step 1: Rewrite the function

The function can be expressed as: $y = \prod_{i=1}^N (x_i + 1)^2.$ This is a product of $N$ terms, where each term is $(x_i + 1)^2$.

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Step 2: Take the natural logarithm (useful for simplifying differentiation)

Using the property of logarithms: $\ln y = \ln \left( \prod_{i=1}^N (x_i + 1)^2 \right) = 2 \sum_{i=1}^N \ln(x_i + 1).$

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Step 3: Differentiate both sides with respect to $x_m$

Using the chain rule: $\frac{1}{y} \frac{\partial y}{\partial x_m} = 2 \sum_{i=1}^N \frac{\partial}{\partial x_m} \ln(x_i + 1).$ The derivative of $\ln(x_i + 1)$ with respect to $x_m$ is zero unless $i = m$, because the terms $x_i + 1$ are independent of $x_m$ for $i \neq m$. Thus:

\begin{cases} \frac{1}{x_m + 1}, & \text{if } i = m, \\ 0, & \text{if } i \neq m. \end{cases}$$ So the sum reduces to: $$\frac{1}{y} \frac{\partial y}{\partial x_m} = \frac{2}{x_m + 1}.$$ --- ### Step 4: Solve for $$\frac{\partial y}{\partial x_m}$$ Multiply through by $$y$$ to isolate $$\frac{\partial y}{\partial x_m}$$: $$\frac{\partial y}{\partial x_m} = y \cdot \frac{2}{x_m + 1}.$$ --- ### Step 5: Substitute $$y$$ back Recall that $$y = \prod_{i=1}^N (x_i + 1)^2$$. Substituting: $$\frac{\partial y}{\partial x_m} = \left( \prod_{i=1}^N (x_i + 1)^2 \right) \cdot \frac{2}{x_m + 1}.$$ --- ### Final Answer $$\boxed{\frac{\partial y}{\partial x_m} = 2 \prod_{i=1}^N (x_i + 1)^2 \cdot \frac{1}{x_m + 1}}.$$ Let me know if you'd like additional details or have further questions! --- ### Relative Questions 1. What is the significance of the chain rule in differentiation? 2. How do logarithmic properties simplify differentiation in problems involving products? 3. Can this result be generalized for different exponents instead of $$2$$? 4. How would you differentiate $$y = \prod_{i=1}^N (x_i + 1)^k$$ for an arbitrary $$k$$? 5. How does this result change if $$x_i + 1$$ is replaced by $$f(x_i)$$? --- ### Tip When dealing with products of terms, taking the natural logarithm often simplifies the differentiation process significantly!