Second Order Partial Derivatives in Calculus

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

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


More on partial derivatives and mutlivariable functions.
Multivariable Functions