Calculate the first and second partial derivatives for the functions f(x,y) = x3 + 5xy + 2y − 2 and f(x,y) = 8 + cos 2x - xy2.

$$f_x(x,y) = \frac{\partial}{\partial x}(x^3 + 5xy + 2y - 2) = 3x^2 + 5y$$ $$f_y(x,y) = \frac{\partial}{\partial y}(x^3 + 5xy + 2y - 2) = 5x + 2$$

$$f_{xx}(x,y) = \frac{\partial}{\partial x}(3x^2 + 5y) = 6x$$ $$f_{xy}(x,y) = \frac{\partial}{\partial y}(3x^2 + 5y) = 5$$ $$f_{yy}(x,y) = \frac{\partial}{\partial y}(5x + 2) = 0$$

The derivatives for $f(x,y) = x^3 + 5xy + 2y - 2$ are:
$f_x(x,y) = \boxed{3x^2 + 5y}$
$f_y(x,y) = \boxed{5x + 2}$
$f_{xx}(x,y) = \boxed{6x}$
$f_{xy}(x,y) = \boxed{5}$
$f_{yy}(x,y) = \boxed{0}$

$$f_x(x,y) = \frac{\partial}{\partial x}(8 + \cos 2x - xy^2) = -2\sin 2x - y^2$$ $$f_y(x,y) = \frac{\partial}{\partial y}(8 + \cos 2x - xy^2) = -2xy$$

$$f_{xx}(x,y) = \frac{\partial}{\partial x}(-2\sin 2x - y^2) = -4\cos 2x$$ $$f_{xy}(x,y) = \frac{\partial}{\partial y}(-2\sin 2x - y^2) = -2y$$ $$f_{yy}(x,y) = \frac{\partial}{\partial y}(-2xy) = -2x$$

The derivatives for $f(x,y) = 8 + \cos 2x - xy^2$ are:
$f_x(x,y) = \boxed{-2\sin 2x - y^2}$
$f_y(x,y) = \boxed{-2xy}$
$f_{xx}(x,y) = \boxed{-4\cos 2x}$
$f_{xy}(x,y) = \boxed{-2y}$
$f_{yy}(x,y) = \boxed{-2x}$