# 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