The given function is $f(x, y) = x^2 e^{-x - 2y}$, and you are asked to compute the value of $f_x(0,0) + f_y(0,0) + f_{yy}(0,0)$.

Step-by-Step Solution:

1. Compute $f_x(x, y)$: $f(x, y) = x^2 e^{-x - 2y}$ To find the partial derivative with respect to $x$, apply the product and chain rules: $f_x(x, y) = 2x \cdot e^{-x - 2y} + x^2 \cdot (-1) e^{-x - 2y}$ Simplifying: $f_x(x, y) = e^{-x - 2y} \left( 2x - x^2 \right)$ Evaluate at $(x, y) = (0, 0)$: $f_x(0, 0) = e^{-(0) - 2(0)} \cdot (2(0) - 0^2) = 0$

2. Compute $f_y(x, y)$: Now, take the partial derivative with respect to $y$: $f_y(x, y) = x^2 \cdot (-2) e^{-x - 2y}$ Simplify: $f_y(x, y) = -2x^2 e^{-x - 2y}$ Evaluate at $(x, y) = (0, 0)$: $f_y(0, 0) = -2(0^2) \cdot e^{-(0) - 2(0)} = 0$

3. Compute $f_{yy}(x, y)$: First, recall that: $f_y(x, y) = -2x^2 e^{-x - 2y}$ Now take the partial derivative of this expression with respect to $y$: $f_{yy}(x, y) = -2x^2 \cdot (-2) e^{-x - 2y}$ Simplify: $f_{yy}(x, y) = 4x^2 e^{-x - 2y}$ Evaluate at $(x, y) = (0, 0)$: $f_{yy}(0, 0) = 4(0^2) e^{-(0) - 2(0)} = 0$

Final Result:

$f_x(0, 0) + f_y(0, 0) + f_{yy}(0, 0) = 0 + 0 + 0 = 0$

Therefore, the value of the expression is 0.

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

Here are some related questions for deeper understanding:

1. How would the result change if the function involved a different exponential term?
2. What are the rules for finding partial derivatives of products involving exponentials?
3. How can you use Taylor series to approximate the value of functions around a point like $(0,0)$?
4. Can you find the second mixed partial derivative $f_{xy}(x,y)$ for this function?
5. What does the result $f_x(0, 0) + f_y(0, 0) + f_{yy}(0, 0) = 0$ indicate about the behavior of the function at the origin?

Tip: When differentiating products, be sure to apply the product rule carefully, especially with functions involving exponentials.