To find the quadratic approximation of the function $f(x, y) = x e^y$ near the origin using Taylor's formula, we'll proceed as follows:

1. Taylor's Formula Overview:

The quadratic approximation of a function $f(x, y)$ around the origin $(0, 0)$ is given by:

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

2. Function Evaluation at the Origin:

First, evaluate the function $f(x, y)$ at the origin:

$f(0, 0) = 0 \cdot e^0 = 0$

3. First Partial Derivatives:

Next, calculate the first partial derivatives of $f(x, y)$:

$\frac{\partial f}{\partial x} = e^y, \quad \frac{\partial f}{\partial y} = x e^y$

Evaluate them at the origin:

$\frac{\partial f}{\partial x}(0, 0) = e^0 = 1, \quad \frac{\partial f}{\partial y}(0, 0) = 0 \cdot e^0 = 0$

4. Second Partial Derivatives:

Now, calculate the second partial derivatives:

$\frac{\partial^2 f}{\partial x^2} = 0, \quad \frac{\partial^2 f}{\partial y^2} = x e^y, \quad \frac{\partial^2 f}{\partial x \partial y} = e^y$

Evaluate them at the origin:

$\frac{\partial^2 f}{\partial x^2}(0, 0) = 0, \quad \frac{\partial^2 f}{\partial y^2}(0, 0) = 0 \cdot e^0 = 0, \quad \frac{\partial^2 f}{\partial x \partial y}(0, 0) = e^0 = 1$

5. Quadratic Approximation:

Substitute all these into the quadratic approximation formula:

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

Simplify the expression:

$f(x, y) \approx x + xy$

Final Answer:

The quadratic approximation of $f(x, y) = x e^y$ near the origin is:

$f(x, y) \approx x + xy$

Would you like more details on any step, or do you have any questions?

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Here are 8 related questions:

1. What is the general form of Taylor series for functions of multiple variables?
2. How does the Taylor series change if we expand around a point other than the origin?
3. Can the quadratic approximation be used to estimate errors?
4. How do the first partial derivatives influence the linear approximation?
5. What happens to the quadratic approximation if $f(x,y)$ has no mixed partial derivative?
6. What is the significance of the second partial derivatives in the quadratic approximation?
7. How does the quadratic approximation differ from the linear approximation?
8. Can we extend this method to higher-order approximations?

Tip: When calculating partial derivatives, ensure that you substitute the values after deriving to avoid unnecessary complications.