Question

Chapter 12.3, Pg 711 Question 10 find \(f_x, \ f_y, \ f_{xx}, \ f_{yy}, \ f_{xy}, \ f_{yx}\)
Where \(f(x,y) = x^2y + 3x^2 + 4y - 5\).

Solution

The partial derivatives are given as:
\[ \begin{split} f_x &= \frac{\partial f}{\partial x} = 2xy + 6 \\ f_y &= \frac{\partial f}{\partial y} = x^2 + 4 \\ f_{xx} &= \frac{\partial^2 f}{\partial x^2} =\frac{\partial }{\partial x} (\frac{\partial f}{\partial x}) = 2y + 6 \\ f_{yy} &= \frac{\partial^2 f}{\partial y^2} =\frac{\partial }{\partial y} (\frac{\partial f}{\partial y}) = 0 \\ f_{xy} &= \frac{\partial^2 f}{\partial x \partial y} =\frac{\partial }{\partial x} (\frac{\partial f}{\partial y}) = 2x \\ f_{yx} &= \frac{\partial^2 f}{\partial y \partial x} =\frac{\partial }{\partial y} (\frac{\partial f}{\partial x}) = 2x \\ \end{split} \] Note: \(f_{xy} = f_{yx} \ \ =>\ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}\)
i.e Mixed partial derivatives are equal.