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The problem is asking to find the second-order mixed partial derivative \(\frac{\partial^2 f}{\partial x \partial y}\) where \( f(x, y) = 2y^2 \) and \( x = 1 + \ln(y) \).

The process involves using the chain rule and the relationship between \( x \) and \( y \) to differentiate.

### Step-by-Step Solution:
1. **Express \( y \) in terms of \( x \):** From \( x = 1 + \ln(y) \), we can express \( y \) as \( y = e^{x-1} \).

2. **Calculate \( \frac{\partial f}{\partial y} \):**
   Given \( f(x, y) = 2y^2 \), first differentiate \( f \) with respect to \( y \):
   \[
   \frac{\partial f}{\partial y} = 4y
   \]

3. **Calculate \( \frac{\partial^2 f}{\partial x \partial y} \):**
   We need to use the chain rule to differentiate \( \frac{\partial f}{\partial y} \) with respect to \( x \). Since \( y \) is a function of \( x \), we have:
   \[
   \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x}(4y) = 4 \cdot \frac{\partial y}{\partial x}
   \]

4. **Determine \( \frac{\partial y}{\partial x} \):**
   From \( y = e^{x-1} \), differentiate \( y \) with respect to \( x \):
   \[
   \frac{\partial y}{\partial x} = e^{x-1} = y
   \]

5. **Substitute \( \frac{\partial y}{\partial x} \) back into the expression for \( \frac{\partial^2 f}{\partial x \partial y} \):**
   \[
   \frac{\partial^2 f}{\partial x \partial y} = 4y \cdot y = 4y^2
   \]

Therefore, the final answer is:
\[
\frac{\partial^2 f}{\partial x \partial y} = 4y^2
\]

### Do you want more details or have any other questions?

Here are five questions that relate to this topic:
1. How does the chain rule apply in multivariable calculus?
2. What is the significance of mixed partial derivatives?
3. Can mixed partial derivatives have different results if the order is changed?
4. What does Clairaut's theorem say about mixed partial derivatives?
5. How does implicit differentiation work in this context?

**Tip:** Always verify if the function is continuously differentiable before assuming mixed partial derivatives are equal.

Learn how to calculate the second-order mixed partial derivative \(\frac{\partial^2 f}{\partial x \partial y}\) for the function \( f(x, y) = 2y^2 \), where \( x = 1 + \ln(y) \). Detailed step-by-step solution using the chain rule and Clairaut's theorem included.

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