Using Taylor’s series expand e
x
log (1 + y) up to the second order terms about (0, 0).

To expand \( e^x \log(1 + y) \) up to the second-order terms around \( (0, 0) \) using a Taylor series, we follow these steps:

### 1. Function Definition
Let \( f(x, y) = e^x \log(1 + y) \).

We want to expand this function around \( (x, y) = (0, 0) \), including terms up to second order.

### 2. Taylor Expansion Formula
The Taylor expansion of a function of two variables \( f(x, y) \) around \( (0, 0) \) is given by:

\[
f(x, y) \approx f(0, 0) + \frac{\partial f}{\partial x}(0, 0) x + \frac{\partial f}{\partial y}(0, 0) y + \frac{1}{2} \left( \frac{\partial^2 f}{\partial x^2}(0, 0) x^2 + 2 \frac{\partial^2 f}{\partial x \partial y}(0, 0) xy + \frac{\partial^2 f}{\partial y^2}(0, 0) y^2 \right)
\]

### 3. Compute Function and Derivatives at \( (0, 0) \)

#### \( f(0, 0) \):
\[
f(0, 0) = e^0 \log(1 + 0) = 1 \cdot 0 = 0
\]

#### First-order partial derivatives:
- \( \frac{\partial f}{\partial x} = e^x \log(1 + y) \), so at \( (0, 0) \):
\[
\frac{\partial f}{\partial x}(0, 0) = e^0 \log(1 + 0) = 1 \cdot 0 = 0
\]

- \( \frac{\partial f}{\partial y} = \frac{e^x}{1 + y} \), so at \( (0, 0) \):
\[
\frac{\partial f}{\partial y}(0, 0) = \frac{e^0}{1 + 0} = 1
\]

#### Second-order partial derivatives:
- \( \frac{\partial^2 f}{\partial x^2} = e^x \log(1 + y) \), so at \( (0, 0) \):
\[
\frac{\partial^2 f}{\partial x^2}(0, 0) = e^0 \log(1 + 0) = 1 \cdot 0 = 0
\]

- \( \frac{\partial^2 f}{\partial y^2} = -\frac{e^x}{(1 + y)^2} \), so at \( (0, 0) \):
\[
\frac{\partial^2 f}{\partial y^2}(0, 0) = -\frac{e^0}{(1 + 0)^2} = -1
\]

- \( \frac{\partial^2 f}{\partial x \partial y} = \frac{e^x}{1 + y} \), so at \( (0, 0) \):
\[
\frac{\partial^2 f}{\partial x \partial y}(0, 0) = \frac{e^0}{1 + 0} = 1
\]

### 4. Combine the Terms
Now, using the Taylor expansion formula, we plug in the values we've computed:

\[
f(x, y) \approx 0 + 0 \cdot x + 1 \cdot y + \frac{1}{2} \left( 0 \cdot x^2 + 2 \cdot 1 \cdot xy + (-1) \cdot y^2 \right)
\]

Simplifying:

\[
f(x, y) \approx y + xy - \frac{1}{2} y^2
\]

### Final Answer:
The second-order Taylor expansion of \( e^x \log(1 + y) \) around \( (0, 0) \) is:

\[
f(x, y) \approx y + xy - \frac{1}{2} y^2
\]

Would you like to explore the details further or ask any questions about the steps?

Here are 5 questions to expand on this topic:
1. How do we generalize Taylor series for higher dimensions?
2. What is the significance of second-order terms in a Taylor expansion?
3. Can you expand \( e^x \log(1 + y) \) to third order?
4. How does the mixed partial derivative \( \frac{\partial^2 f}{\partial x \partial y} \) contribute to the expansion?
5. Why is the expansion centered around \( (0, 0) \) instead of another point?

**Tip:** When expanding around a point, ensure you calculate all partial derivatives correctly up to the desired order.

Using Taylor’s series expand e^x log(1 + y) up to the second order terms about (0, 0).

Learn how to expand e^x log(1 + y) using a Taylor series up to second-order terms around (0, 0). This solution involves partial derivatives and covers key concepts in multivariable calculus. Suitable for university-level students studying advanced calculus.

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