For a two variable function f(x , y), we can define 4 second order partial derivatives along with their notations.

\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x } \left (\dfrac{\partial f}{\partial x} \right ) = \frac{\partial}{\partial x } (f_x) = (f_x)_x = f_{xx}

\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y } \left (\dfrac{\partial f}{\partial y} \right ) = \frac{\partial}{\partial y } (f_y) = (f_y)_y = f_{yy}

\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y } \left (\dfrac{\partial f}{\partial x} \right ) = \frac{\partial}{\partial y } (f_x) = (f_x)_y = f_{xy}

\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x } \left (\dfrac{\partial f}{\partial y} \right ) = \frac{\partial}{\partial x } (f_y) = (f_y)_x = f_{yx}

__Example 1:__ Find f_{xx}, f_{yy} given that f(x , y) = sin (x y)

__solution to Example 1:__

f_{xx} may be calculated as follows

f_{xx} = ∂^{2}f / ∂x^{2} = ∂(∂f / ∂x) / ∂x

= ∂(∂[ sin (x y) ]/ ∂x) / ∂x

= ∂(y cos (x y) ) / ∂x

= - y^{2} sin (x y) )

f_{yy} can be calculated as follows

f_{yy} = ∂^{2}f / ∂y^{2} = ∂(∂f / ∂y) / ∂y

= ∂(∂[ sin (x y) ]/ ∂y) / ∂y

= ∂(x cos (x y) ) / ∂y

= - x^{2} sin (x y) )

__Example 2:__ Find f_{xx}, f_{yy}, f_{xy}, f_{yx} given that f(x , y) = x^{3} + 2 x y.

__solution to Example 2:__

f_{xx} is calculated as follows

f_{xx} = ∂^{2}f / ∂x^{2} = ∂(∂f / ∂x) / ∂x

= ∂(∂[ x^{3} + 2 x y ]/ ∂x) / ∂x

= ∂( 3 x^{2} + 2 y ) / ∂x

= 6x

f_{yy} is calculated as follows

f_{yy} = ∂^{2}f / ∂y^{2} = ∂(∂f / ∂y) / ∂y

= ∂(∂[ x^{3} + 2 x y ]/ ∂y) / ∂y

= ∂( 2x ) / ∂y

= 0

f_{xy} is calculated as follows

f_{xy} = ∂^{2}f / ∂y∂x = ∂(∂f / ∂x) / ∂y

= ∂(∂[ x^{3} + 2 x y ]/ ∂x) / ∂y

= ∂( 3 x^{2} + 2 y ) / ∂y

= 2

f_{yx} is calculated as follows

f_{yx} = ∂^{2}f / ∂x∂y = ∂(∂f / ∂y) / ∂x

= ∂(∂[ x^{3} + 2 x y ]/ ∂y) / ∂x

= ∂( 2x ) / ∂x

= 2

__Example 3:__ Find f_{xx}, f_{yy}, f_{xy}, f_{yx} given that f(x , y) = x^{3}y^{4} + x^{2} y.

__solution to Example 2:__

f_{xx} is calculated as follows

f_{xx} = ∂^{2}f / ∂x^{2} = ∂(∂f / ∂x) / ∂x

= ∂(∂[ x^{3}y^{4} + x^{2} y ]/ ∂x) / ∂x

= ∂( 3 x^{2}y^{4} + 2 x y) / ∂x

= 6x y^{4} + 2y

f_{yy} is calculated as follows

f_{yy} = ∂^{2}f / ∂y^{2} = ∂(∂f / ∂y) / ∂y

= ∂(∂[ x^{3}y^{4} + x^{2} y ]/ ∂y) / ∂y

= ∂( 4 x^{3}y^{3} + x^{2} ) / ∂y

= 12 x^{3}y^{2}

f_{xy} is calculated as follows

f_{xy} = ∂^{2}f / ∂y∂x = ∂(∂f / ∂x) / ∂y

= ∂(∂[ x^{3}y^{4} + x^{2} y ]/ ∂x) / ∂y

= ∂( 3 x^{2}y^{4} + 2 x y ) / ∂y

= 12 x^{2}y^{3} + 2 x

f_{yx} is calculated as follows

f_{yx} = ∂^{2}f / ∂x∂y = ∂(∂f / ∂y) / ∂x

= ∂(∂[ x^{3}y^{4} + x^{2} y ]/ ∂y) / ∂x

= ∂(4 x^{3}y^{3} + x^{2}) / ∂x

= 12 x^{2}y^{3} + 2x

More on partial derivatives and mutlivariable functions.
Multivariable Functions