Second-order partial derivatives

For a function of one variable, differentiating $f'$ again gives the second derivative $f''$. A function of two variables, $f(x, y)$, has two first partials, $f_x$ and $f_y$. Each is itself a function of $x$ and $y$, so each can be differentiated by $x$ or by $y$. That gives four second-order partial derivatives, each defined only where both of its steps exist.

Each of the four has a subscript form and a $\partial$ form:

The notation extends to more variables. A function of three variables has first partials $f_x$, $f_y$ and $f_z$, and $f_{xz} = \left(f_x\right)_z$ is the partial of $f_x$ with respect to $z$.

For a function $f(x,y)$ of two variables, the graph $z=f(x,y)$ is a surface. The two second-order partials that repeat a variable, $f_{xx}$ and $f_{yy}$, each describe one cross-section of that surface.

Each is the second derivative of a curve, so its sign gives the concavity of that cross-section.

Toggle the two views below to see the two cross-sections of one surface through the same point, first with $y$ held fixed and then with $x$ held fixed.

Nothing here is new. Each of $f_{xx}$ and $f_{yy}$ is the second derivative we already know, applied to a cross-section of the surface.

For $f(x,y) = x^3 + xy - y^4$, a calculation at the point $(1,2)$ gives

Using these values, and without differentiating:

1. Name the cross-section that $f_{xx}(1,2)$ describes, and state its concavity there.
2. Name the cross-section that $f_{yy}(1,2)$ describes, and state its concavity there.

Solution

1. Both subscripts of $f_{xx}$ are $x$, so we hold $y = 2$ fixed at both steps. That leaves the cross-section $z = f(x,2)$, a function of $x$ alone, and $f_{xx}(1,2)$ is its ordinary second derivative at $x = 1$. That value is positive, so this cross-section is concave up at $x = 1$.

2. For $f_{yy}$ we hold $x = 1$ fixed instead, which leaves the cross-section $z = f(1,y)$, a function of $y$ alone, and $f_{yy}(1,2)$ is its ordinary second derivative at $y = 2$. That value is negative, so this cross-section is concave down at $y = 2$.

The two remaining second-order partials are the mixed partials, $f_{xy}$ and $f_{yx}$. Neither is the second derivative of a single cross-section, because the two steps use different variables. Instead, each compares one cross-section with its neighbours.

$f_{yx}(a,b)$ measures that change. Since $f_{yx} = \left(f_y\right)_x$, it differentiates the slope in the $y$ direction, $f_y$, with respect to $x$.

Slide the cross-section below to watch the slope in the $y$ direction change with $x$.

As with any derivative, the sign tells us the direction of the change. A positive $f_{yx}(a,b)$ means the slope in the $y$ direction is increasing as $x$ increases, and a negative one means it is decreasing. A value of $0$ means that, at $(a,b)$, moving in the $x$ direction is not changing the slope in the $y$ direction.

For a function $f(x,y)$, a calculation at the point $(3,1)$ gives

Using these values, and without differentiating:

1. State which first partial $f_{yx}(3,1)$ tracks, and in which direction it tracks the change.
2. Describe what the two values say about the surface $z = f(x,y)$ at $(3,1)$.

Solution

1. Subscripts are read left to right, so $f_{yx} = \left(f_y\right)_x$: the first partial is $f_y$, and it is differentiated by $x$. So $f_{yx}(3,1)$ is the rate at which $f_y$ changes as $x$ increases, with $y$ held fixed at $1$.

2. At $(3,1)$ the slope in the $y$ direction is $f_y(3,1) = 2$, so the cross-section $z = f(3,y)$ is rising at $y = 1$. The value $f_{yx}(3,1) = -5$ is negative, so that slope is decreasing as $x$ increases. A small increase in $x$ brings us to a cross-section that is less steep in the $y$ direction at $y = 1$.