Learning Outcomes
- Determine the higher-order derivatives of a function of two variables.
Consider the function
[latex]\large{f\,(x,\ y)=2x^{3}-4xy^{2}+5y^{3}-6xy+5x-4y+12}.[/latex]
Its partial derivatives are
[latex]\large{\frac{\partial f}{\partial x}=6x^{2}-4y^{2}-6y+5}[/latex] and [latex]\large{\frac{\partial f}{\partial y}=-8xy+15y^{2}-6x-4}.[/latex]
Each of these partial derivatives is a function of two variables, so we can calculate partial derivatives of these functions. Just as with derivatives of single-variable functions, we can call these second-order derivatives, third-order derivatives, and so on. In general, they are referred to as higher-order partial derivatives. There are four second-order partial derivatives for any function (provided they all exist):
[latex]\large{\frac{\partial^{2}f}{\partial x^{2}}=\frac{\partial}{\partial x}\left[\frac{\partial f}{\partial x}\right],\ \frac{\partial^{2}f}{\partial x\partial y}=\frac{\partial}{\partial x}\left[\frac{\partial f}{\partial y}\right],\ \frac{\partial^{2}f}{\partial y\partial x}=\frac{\partial}{\partial y}\left[\frac{\partial f}{\partial x}\right],\ \frac{\partial^{2}f}{\partial y^{2}}=\frac{\partial}{\partial y}\left[\frac{\partial f}{\partial y}\right]}.[/latex]
An alternative notation for each is [latex]f_{xx},\ f_{yx},\ f_{xy}[/latex], and [latex]f_{yy}[/latex], respectively. Higher-order partial derivatives calculated with respect to different variables, such as [latex]f_{xy}[/latex] and [latex]f_{yx}[/latex], are commonly called mixed partial derivatives.
Example: Calculating Second Partial Derivatives
Calculate all four second partial derivates for the function
[latex]\large{f\,(x,\ y)=xe^{-3y}+\sin{(2x-5y)}}[/latex].
Show Solution
To calculate [latex]\partial^{2}f/\partial x^{2}[/latex] and [latex]\partial^{2}f/\partial y\partial x[/latex], we first calculate [latex]\partial f/\partial x[/latex]:
[latex]\large{\frac{\partial f}{\partial x}=e^{-3y}+2\cos{(2x-5y)}}.[/latex]
To calculate [latex]\partial^{2}f/\partial x^{2}[/latex], differentiate [latex]\partial f/\partial x[/latex] with respect to [latex]x[/latex]:
[latex]\begin{array}{ccc}\hfill {\frac{{\partial}^{2} f}{\partial {x^{2}}}} & =\hfill & {\frac{\partial}{\partial x}\left[\frac{\partial f}{\partial x}\right]} \hfill \\ \hfill & =\hfill & {\frac{\partial}{\partial x}\left[e^{-3y}+2\cos{(2x-5y)}\right]} \hfill \\ \hfill & =\hfill & {-4\sin{(2x-5y)}.} \hfill \\ \hfill \end{array}[/latex]
To calculate [latex]\partial^{2}f/\partial y\partial x[/latex], differentiate [latex]\partial f/\partial x[/latex] with respect to [latex]y[/latex]:
[latex]\begin{array}{ccc}\hfill {\frac{\partial^{2}f}{\partial y\partial x}} & =\hfill & {\frac{\partial}{\partial y}\left[\frac{\partial f}{\partial x}\right]} \hfill \\ \hfill & =\hfill & {\frac{\partial}{\partial y}\left[e^{-3y}+2\cos{(2x-5y)}\right]} \hfill \\ \hfill & =\hfill & {-3e^{-3y}+10\sin{(2x-5y)}.} \hfill \\ \hfill \end{array}[/latex]
To calculate [latex]\partial^{2}f/\partial x\partial y[/latex] and [latex]\partial^{2}f/\partial y^{2}[/latex], first calculate [latex]\partial f/\partial y[/latex]:
[latex]\large{\frac{\partial f}{\partial y}=-3xe^{-3y}-5\cos{(2x-5y)}.}[/latex]
To calculate [latex]\partial^{2}f/\partial x\partial y[/latex], differentiate [latex]\partial f/\partial y[/latex] with respect to [latex]x[/latex]:
[latex]\begin{array}{ccc}\hfill {\frac{\partial^{2}f}{\partial x\partial y}} & =\hfill & {\frac{\partial}{\partial x}\left[\frac{\partial f}{\partial y}\right]} \hfill \\ \hfill & =\hfill & {\frac{\partial}{\partial x}\left[-3xe^{-3y}-5\cos{(2x-5y)}\right]} \hfill \\ \hfill & =\hfill & {-3e^{-3y}+10\sin{(2x-5y)}.} \hfill \\ \hfill \end{array}[/latex]
To calculate [latex]\partial^{2}f/\partial^{2} y[/latex], differentiate [latex]\partial f/\partial y[/latex] with respect to [latex]y[/latex]:
[latex]\begin{array}{ccc}\hfill {\frac{\partial^{2}f}{\partial y}} & =\hfill & {\frac{\partial}{\partial y}\left[\frac{\partial f}{\partial y}\right]} \hfill \\ \hfill & =\hfill & {\frac{\partial}{\partial y}\left[-3xe^{-3y}-5\cos{(2x-5y)}\right]} \hfill \\ \hfill & =\hfill & {9xe^{-3y}-25\sin{(2x-5y)}.} \hfill \\ \hfill \end{array}[/latex]
TRY IT
Calculate all four second partial derivatives for the function
[latex]\large{f\,(x,\ y)=\sin{(3x-2y)}+\cos{(x+4y)}.}[/latex]
Show Solution
[latex]\frac{\partial^{2}f}{\partial x^{2}}=-9\sin{(3x-2y)}-\cos{(x+4y)}[/latex]
[latex]\frac{\partial^{2}f}{\partial x\partial y}=6\sin{(3x-2y)}-4\cos{(x+4y)}[/latex]
[latex]\frac{\partial^{2}f}{\partial y\partial x}=6\sin{(3x-2y)}-4\cos{(x+4y)}[/latex]
[latex]\frac{\partial^{2}f}{\partial y^{2}}=-4\sin{(3x-2y)}-16\cos{(x+4y)}[/latex]
Watch the following video to see the worked solution to the above Try It
You can view the transcript for “CP 4.17” here (opens in new window).At this point we should notice that, in both the previous example and the “Try It,” it was true that [latex]\partial^{2}f/\partial x\partial y=\partial^{2}f/\partial y\partial x[/latex]. Under certain conditions, this is always true. In fact, it is a direct consequence of the following theorem.
Equality of Mixed PArtial Derivatives (Clairaut’s Theorem)
Suppose that [latex]f\,(x,\ y)[/latex] is defined on an open disk [latex]D[/latex] that contains the point [latex](a,\ b)[/latex]. If the functions [latex]f_{xy}[/latex] and [latex]f_{yx}[/latex] are continuous on [latex]D[/latex], then [latex]f_{xy}=f_{yx}[/latex].
Clairaut’s theorem guarantees that as long as mixed second-order derivatives are continuous, the order in which we choose to differentiate the functions (i.e., which variable goes first, then second, and so on) does not matter. It can be extended to higher-order derivatives as well. The proof of Clairaut’s theorem can be found in most advanced calculus books.
Two other second-order partial derivatives can be calculated for any function [latex]f\,(x,\ y)[/latex]. The partial derivative [latex]f_{xx}[/latex] is equal to the partial derivative of [latex]f_x[/latex] with respect to [latex]x[/latex], and [latex]f_{yy}[/latex] is equal to the partial derivative of [latex]f_y[/latex] with respect to [latex]y[/latex].
