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.
second order partial derivatives

Examples with Detailed Solutions

Example 1

Find fxx, fyy given that f(x , y) = sin (x y)
solution to Example 1:
fxx may be calculated as follows
f
xx = ∂2f / ∂x2 = ∂(∂f / ∂x) / ∂x
= ∂(∂[ sin (x y) ]/ ∂x) / ∂x
= ∂(y cos (x y) ) / ∂x
= - y
2 sin (x y) )
fyy can be calculated as follows
f
yy = ∂2f / ∂y2 = ∂(∂f / ∂y) / ∂y
= ∂(∂[ sin (x y) ]/ ∂y) / ∂y
= ∂(x cos (x y) ) / ∂y
= - x
2 sin (x y) )

Example 2

Find fxx, fyy, fxy, fyx given that f(x , y) = x3 + 2 x y.
solution to Example 2:
fxx is calculated as follows
f
xx = ∂2f / ∂x2 = ∂(∂f / ∂x) / ∂x
= ∂(∂[ x
3 + 2 x y ]/ ∂x) / ∂x
= ∂( 3 x
2 + 2 y ) / ∂x
= 6x
fyy is calculated as follows
f
yy = ∂2f / ∂y2 = ∂(∂f / ∂y) / ∂y
= ∂(∂[ x
3 + 2 x y ]/ ∂y) / ∂y
= ∂( 2x ) / ∂y
= 0
fxy is calculated as follows
f
xy = ∂2f / ∂y∂x = ∂(∂f / ∂x) / ∂y
= ∂(∂[ x
3 + 2 x y ]/ ∂x) / ∂y
= ∂( 3 x
2 + 2 y ) / ∂y
= 2
fyx is calculated as follows
f
yx = ∂2f / ∂x∂y = ∂(∂f / ∂y) / ∂x
= ∂(∂[ x
3 + 2 x y ]/ ∂y) / ∂x
= ∂( 2x ) / ∂x
= 2

Example 3

Find fxx, fyy, fxy, fyx given that f(x , y) = x3y4 + x2 y.
solution to Example 2:
fxx is calculated as follows
f
xx = ∂2f / ∂x2 = ∂(∂f / ∂x) / ∂x
= ∂(∂[ x
3y4 + x2 y ]/ ∂x) / ∂x
= ∂( 3 x
2y4 + 2 x y) / ∂x
= 6x y
4 + 2y
fyy is calculated as follows
f
yy = ∂2f / ∂y2 = ∂(∂f / ∂y) / ∂y
= ∂(∂[ x
3y4 + x2 y ]/ ∂y) / ∂y
= ∂( 4 x
3y3 + x2 ) / ∂y
= 12 x
3y2
fxy is calculated as follows
f
xy = ∂2f / ∂y∂x = ∂(∂f / ∂x) / ∂y
= ∂(∂[ x
3y4 + x2 y ]/ ∂x) / ∂y
= ∂( 3 x
2y4 + 2 x y ) / ∂y
= 12 x
2y3 + 2 x
fyx is calculated as follows
f
yx = ∂2f / ∂x∂y = ∂(∂f / ∂y) / ∂x
= ∂(∂[ x
3y4 + x2 y ]/ ∂y) / ∂x
= ∂(4 x
3y3 + x2) / ∂x
= 12 x
2y3 + 2x
More on partial derivatives and mutlivariable functions.
Multivariable Functions