Introduction to Partial Derivatives: Learn It 3

Giselle Miller

Introduction to Partial Derivatives: Learn It 3

Higher-Order Partial Derivatives

Consider the function

[latex]f(x,y)=2x^{3}-4xy^{2}+5y^{3}-6xy+5x-4y+12[/latex].

Its partial derivatives are

[latex]\dfrac{\partial f}{\partial x}=6x^{2}-4y^{2}-6y+5[/latex] and [latex]\dfrac{\partial f}{\partial y}=-8xy+15y^{2}-6x-4[/latex].

Each of these partial derivatives is itself a function of two variables, so we can calculate partial derivatives of these functions. Just as with single-variable derivatives, we call these second-order derivatives, third-order derivatives, and so on. In general, they are 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 these is [latex]f_{xx}[/latex], [latex]f_{yx}[/latex], [latex]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 called mixed partial derivatives.

In [latex]\frac{\partial^{2}f}{\partial y\partial x}[/latex], read the variables from right to left: differentiate first with respect to [latex]x[/latex], then with respect to [latex]y[/latex].

In subscript notation [latex]f_{xy}[/latex], read left to right: differentiate first with respect to [latex]x[/latex], then with respect to [latex]y[/latex].

Calculate all four second partial derivates for the function

[latex]\large{f\,(x,\ y)=xe^{-3y}+\sin{(2x-5y)}}[/latex].

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]

Notice something interesting in the previous example: [latex]\dfrac{\partial^{2}f}{\partial x\partial y}=\dfrac{\partial^{2}f}{\partial y\partial x}[/latex]. Both mixed partial derivatives equal [latex]-3e^{-3y}+10\sin(2x-5y)[/latex]. Is this always true? Under certain conditions, yes. In fact, this is guaranteed by the following theorem.

equality of mixed partial derivatives (Clairaut’s Theorem)

Suppose [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 differentiate doesn’t matter. We can differentiate with respect to [latex]x[/latex] first then [latex]y[/latex], or [latex]y[/latex] first then [latex]x[/latex]—we’ll get the same result. This theorem can be extended to higher-order derivatives as well.

Beyond the mixed partial derivatives, we also have [latex]f_{xx}[/latex] (the partial derivative of [latex]f_x[/latex] with respect to [latex]x[/latex]) and [latex]f_{yy}[/latex] (the partial derivative of [latex]f_y[/latex] with respect to [latex]y[/latex]). These aren’t mixed—they differentiate with respect to the same variable twice.