2 Second partial derivatives

Performing two successive partial differentiations of $f\left(x,y\right)$ with respect to $x$ (holding $y$ constant) is denoted by $\frac{{\partial }^{2}f}{\partial x^2}$ (or $f_{xx}\left(x,y\right)$ ) and is defined by

$\frac{{\partial }^{2}f}{\partial x^2}\equiv \frac{\partial }{\partial x}\left(\frac{\partial f}{\partial x}\right)$

For functions of two or more variables as well as $\frac{{\partial }^{2}f}{\partial x^2}$ other second-order partial derivatives can be obtained. Most obvious is the second derivative of $f\left(x,y\right)$ with respect to $y$ is denoted by $\frac{{\partial }^{2}f}{\partial y^2}$ (or $f_{yy}\left(x,y\right)$ ) which is defined as:

$\frac{{\partial }^{2}f}{\partial y^2}\equiv \frac{\partial }{\partial y}\left(\frac{\partial f}{\partial y}\right)$

Example 5

Find $\frac{{\partial }^{2}f}{\partial x^2}$ and $\frac{{\partial }^{2}f}{\partial y^2}$ for $f\left(x,y\right)=x^3+x^2{y}^2+2y^3+2x+y$ .

Solution

$\frac{\partial f}{\partial x}=3x^2+2x{y}^2+0+2+0=3x^2+2x{y}^2+2$

$\frac{{\partial }^{2}f}{\partial x^2}\equiv \frac{\partial }{\partial x}\left(\frac{\partial f}{\partial x}\right)=6x+2y^2+0=6x+2y^2.$

$\frac{\partial f}{\partial y}=0+x^2\times 2y+6y^2+0+1=2x^{2}y+6y^2+1$

$\frac{{\partial }^{2}f}{\partial y^2}=\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial y}\right)=2x^2+12y$ .

We can use the alternative notation when evaluating derivatives.

Example 6

Find $f_{xx}\left(-1,1\right)$ and $f_{yy}\left(2,-2\right)$ for $f\left(x,y\right)=x^3+x^2{y}^2+2y^3+2x+y$ .

Solution

$f_{xx}\left(-1,1\right)=6\times \left(-1\right)+2\times {\left(-1\right)}^2=-4.$

$f_{yy}\left(2,-2\right)=2\times {\left(2\right)}^2+12\times \left(-2\right)=-16$

2.1 Mixed second derivatives

It is possible to carry out a partial differentiation of $f\left(x,y\right)$ with respect to $x$ followed by a partial differentiation with respect to $y$ (or vice-versa). The results are examples of mixed derivatives . We must be careful with the notation here.

We use $\frac{{\partial }^{2}f}{\partial x\partial y}$ to mean ‘differentiate first with respect to $y$ and then with respect to $x$ ’ and we use $\frac{{\partial }^{2}f}{\partial y\partial x}$ to mean ‘differentiate first with respect to $x$ and then with respect to $y$ ’:

$\text{i.e.}\frac{{\partial }^{2}f}{\partial x\partial y}\equiv \frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y}\right)\text{and}\frac{{\partial }^{2}f}{\partial y\partial x}\equiv \frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\right).$

(This explains why the order is opposite of what we expect - the derivative ‘operates on the left’.)

Example 7

For $f\left(x,y\right)=x^3+2x^2{y}^2+y^3$ find $\frac{{\partial }^{2}f}{\partial x\partial y}.$

Solution

$\frac{\partial f}{\partial y}=4x^{2}y+3y^2;\frac{{\partial }^{2}f}{\partial x\partial y}=8xy$

The remaining possibility is to differentiate first with respect to $x$ and then with respect to $y$ i.e. $\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\right)$ .

For the function in Example 7 $\frac{\partial f}{\partial x}=3x^2+4x{y}^2$   and $\frac{{\partial }^{2}f}{\partial y\partial x}=8xy.$ Notice that for this function

$\frac{{\partial }^{2}f}{\partial x\partial y}\equiv \frac{{\partial }^{2}f}{\partial y\partial x}.$

This equality of mixed derivatives is true for all functions which you are likely to meet in your studies.

To evaluate a mixed derivative we can use the alternative notation. To evaluate $\frac{{\partial }^{2}f}{\partial x\partial y}$ we write $f_{yx}\left(x,y\right)$ to indicate that the first differentiation is with respect to $y$ . Similarly, $\frac{{\partial }^{2}f}{\partial y\partial x}$ is denoted by $f_{xy}\left(x,y\right)$ .

Example 8

Find $f_{yx}\left(1,2\right)$ for the function $f\left(x,y\right)=x^3+2x^2{y}^2+y^3$

Solution

$f_x=3x^2+4x{y}^2$  and $f_{yx}=8xy\text{so}{f}_{yx}\left(1,2\right)=8\times 1\times 2=16.$

Task!

Find $f_{xx}\left(1,2\right),f_{yy}\left(-2,-1\right),$ $f_{xy}\left(3,3\right)$ for   $f\left(x,y\right)\equiv x^3+3x^2{y}^2+y^2.$

Answer