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
fxx 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) )
fyy 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 fxx, fyy, fxy, fyx given that f(x , y) = x3 + 2 x y.
Solution
fxx is calculated as follows
fxx = ∂2f / ∂x2 = ∂(∂f / ∂x) / ∂x
= ∂(∂[ x3 + 2 x y ]/ ∂x) / ∂x
= ∂( 3 x2 + 2 y ) / ∂x
= 6x
fyy is calculated as follows
fyy = ∂2f / ∂y2 = ∂(∂f / ∂y) / ∂y
= ∂(∂[ x3 + 2 x y ]/ ∂y) / ∂y
= ∂( 2x ) / ∂y
= 0
fxy is calculated as follows
fxy = ∂2f / ∂y∂x = ∂(∂f / ∂x) / ∂y
= ∂(∂[ x3 + 2 x y ]/ ∂x) / ∂y
= ∂( 3 x2 + 2 y ) / ∂y
= 2
fyx is calculated as follows
fyx = ∂2f / ∂x∂y = ∂(∂f / ∂y) / ∂x
= ∂(∂[ x3 + 2 x y ]/ ∂y) / ∂x
= ∂( 2x ) / ∂x
= 2
Example 3
Find fxx, fyy, fxy, fyx given that f(x , y) = x3y4 + x2 y.
Solution
fxx is calculated as follows
fxx = ∂2f / ∂x2 = ∂(∂f / ∂x) / ∂x
= ∂(∂[ x3y4 + x2 y ]/ ∂x) / ∂x
= ∂( 3 x2y4 + 2 x y) / ∂x
= 6x y4 + 2y
fyy is calculated as follows
fyy = ∂2f / ∂y2 = ∂(∂f / ∂y) / ∂y
= ∂(∂[ x3y4 + x2 y ]/ ∂y) / ∂y
= ∂( 4 x3y3 + x2 ) / ∂y
= 12 x3y2
fxy is calculated as follows
fxy = ∂2f / ∂y∂x = ∂(∂f / ∂x) / ∂y
= ∂(∂[ x3y4 + x2 y ]/ ∂x) / ∂y
= ∂( 3 x2y4 + 2 x y ) / ∂y
= 12 x2y3 + 2 x
fyx is calculated as follows
fyx = ∂2f / ∂x∂y = ∂(∂f / ∂y) / ∂x
= ∂(∂[ x3y4 + x2 y ]/ ∂y) / ∂x
= ∂(4 x3y3 + x2) / ∂x
= 12 x2y3 + 2x
