1. **Problem statement:** Expand the function $f(x,y) = e^{x+y}$ using Taylor series about the point $(0,0)$ up to second degree terms. 2. **Formula:** The Taylor series expansion of a function $f(x,y)$ about $(a,b)$ is given by: $$ \begin{aligned} f(x,y) \approx & f(a,b) + f_x(a,b)(x - a) + f_y(a,b)(y - b) \\ & + \frac{1}{2} f_{xx}(a,b)(x - a)^2 + f_{xy}(a,b)(x - a)(y - b) + \frac{1}{2} f_{yy}(a,b)(y - b)^2 \end{aligned} $$ where $f_x$, $f_y$ are first partial derivatives and $f_{xx}$, $f_{xy}$, $f_{yy}$ are second partial derivatives. 3. **Evaluate function and derivatives at $(0,0)$:** - $f(0,0) = e^{0+0} = 1$ - $f_x = \frac{\partial}{\partial x} e^{x+y} = e^{x+y}$, so $f_x(0,0) = 1$ - $f_y = \frac{\partial}{\partial y} e^{x+y} = e^{x+y}$, so $f_y(0,0) = 1$ - $f_{xx} = \frac{\partial^2}{\partial x^2} e^{x+y} = e^{x+y}$, so $f_{xx}(0,0) = 1$ - $f_{xy} = \frac{\partial^2}{\partial x \partial y} e^{x+y} = e^{x+y}$, so $f_{xy}(0,0) = 1$ - $f_{yy} = \frac{\partial^2}{\partial y^2} e^{x+y} = e^{x+y}$, so $f_{yy}(0,0) = 1$ 4. **Substitute into Taylor series:** $$ \begin{aligned} f(x,y) \approx & 1 + 1 \cdot x + 1 \cdot y + \frac{1}{2} \cdot 1 \cdot x^2 + 1 \cdot x y + \frac{1}{2} \cdot 1 \cdot y^2 \\ = & 1 + x + y + \frac{x^2}{2} + x y + \frac{y^2}{2} \end{aligned} $$ 5. **Final answer:** $$ \boxed{f(x,y) \approx 1 + x + y + \frac{x^2}{2} + x y + \frac{y^2}{2}} $$