# Second Order Partial Derivatives in Calculus

Examples with detailed solutions on how to calculate second order partial derivatives are presented.

## Definitions and Notations of Second Order Partial Derivatives

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

## Examples with Detailed Solutions on Second Order Partial Derivatives

Example 1Find f

_{xx}, f

_{yy}given that f(x , y) = sin (x y)

Solution

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

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

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