1. The problem is to find the second-order Taylor expansion of the function $$f(x,y) = e^x \cos y$$ about the point $$(0,0)$$. 2. The Taylor expansion formula for a function of two variables about $$(a,b)$$ up to second order is: $$ f(x,y) \approx f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b) + \frac{1}{2} f_{xx}(a,b)(x-a)^2 + f_{xy}(a,b)(x-a)(y-b) + \frac{1}{2} f_{yy}(a,b)(y-b)^2 $$ where $f_x, f_y$ are first partial derivatives and $f_{xx}, f_{xy}, f_{yy}$ are second partial derivatives. 3. Calculate the function value at $$(0,0)$$: $$ f(0,0) = e^0 \cos 0 = 1 \times 1 = 1 $$ 4. Calculate first partial derivatives: $$ f_x = \frac{\partial}{\partial x} (e^x \cos y) = e^x \cos y $$ $$ f_y = \frac{\partial}{\partial y} (e^x \cos y) = -e^x \sin y $$ Evaluate at $$(0,0)$$: $$ f_x(0,0) = e^0 \cos 0 = 1 $$ $$ f_y(0,0) = -e^0 \sin 0 = 0 $$ 5. Calculate second partial derivatives: $$ f_{xx} = \frac{\partial}{\partial x} (f_x) = \frac{\partial}{\partial x} (e^x \cos y) = e^x \cos y $$ $$ f_{xy} = \frac{\partial}{\partial y} (f_x) = \frac{\partial}{\partial y} (e^x \cos y) = -e^x \sin y $$ $$ f_{yy} = \frac{\partial}{\partial y} (f_y) = \frac{\partial}{\partial y} (-e^x \sin y) = -e^x \cos y $$ Evaluate at $$(0,0)$$: $$ f_{xx}(0,0) = e^0 \cos 0 = 1 $$ $$ f_{xy}(0,0) = -e^0 \sin 0 = 0 $$ $$ f_{yy}(0,0) = -e^0 \cos 0 = -1 $$ 6. Substitute all values into the Taylor expansion formula: $$ f(x,y) \approx 1 + 1 \cdot x + 0 \cdot y + \frac{1}{2} \cdot 1 \cdot x^2 + 0 \cdot xy + \frac{1}{2} \cdot (-1) \cdot y^2 = 1 + x + \frac{x^2}{2} - \frac{y^2}{2} $$ Final answer: $$ f(x,y) \approx 1 + x + \frac{x^2}{2} - \frac{y^2}{2} $$