Partial Derivatives, multivariable and vector calculus lesson

Stand on a hillside and face due east. The ground in front of you has a slope. Turn to face due north and the slope is different. A function of two variables has no single rate of change at a point, but if you move along one coordinate direction and hold the other fixed, you are back to an ordinary derivative. That derivative is a partial derivative.

Key terms

Varying one coordinate

Definition Let $f$ be a function of $x$ and $y$ and let $(a, b)$ be an interior point of its domain. The partial derivative of $f$ with respect to $x$ at $(a, b)$ is

f_x(a, b) = \lim_{h \to 0} \frac{f(a + h, b) - f(a, b)}{h}

and the partial derivative with respect to $y$ is

f_y(a, b) = \lim_{h \to 0} \frac{f(a, b + h) - f(a, b)}{h}

provided these limits exist. For three or more variables, change one coordinate by $h$ and keep all the others fixed. In practice $f_x$ is computed by treating $y$ as a constant and differentiating in $x$ with the one-variable rules. Existence of both partial derivatives at a point says nothing about other directions, and it does not even make $f$ continuous there.

In plain words Freeze every input but one, then take the ordinary derivative in the one that is still moving.

Example For $f(x, y) = 3x^2 y$ at $(1, 3)$ , only $x$ changes in the quotient,

\begin{aligned} &\frac{f(1 + h, 3) - f(1, 3)}{h} \\[0.4em] &\quad = \frac{9(1 + h)^2 - 9}{h} = 18 + 9h, \end{aligned}

so $f_x(1, 3) = 18$ . The shortcut gives the same answer. With $y$ held constant, $f_x = 6xy$ , and at $(1, 3)$ that is $18$ .

Slopes of traces

Definition Let $f_x(a, b)$ exist. The trace of the graph $z = f(x, y)$ in the vertical plane $y = b$ is the curve $z = f(x, b)$ , and $f_x(a, b)$ is the slope of its tangent line at $x = a$ , measured in the $x$ direction. Likewise $f_y(a, b)$ is the slope at $y = b$ of the trace $z = f(a, y)$ in the plane $x = a$ . Both are rates of change of the output per unit change of one input.

In plain words Slice the surface parallel to one axis. The partial derivative is the steepness of the cut, read off like any single-variable slope.

Example For $f(x, y) = 4 - x^2 - 2y^2$ at $(1, 1)$ the height is $f(1, 1) = 1$ , and

f_x = -2x, \qquad f_y = -4y.

So $f_x(1, 1) = -2$ and $f_y(1, 1) = -4$ . Walking in the positive $x$ direction from the point $(1, 1, 1)$ the surface drops $2$ units per unit step, and in the positive $y$ direction it drops $4$ . The tangent line to the trace in the plane $y = 1$ is $z = 1 - 2(x - 1)$ .

Partial derivative notation

Definition For $z = f(x, y)$ the following all denote the partial derivative with respect to $x$ ,

f_x = \frac{\partial f}{\partial x} = \frac{\partial z}{\partial x} = z_x = D_1 f,

and similarly for $y$ , with $D_2 f$ for the second input. The curly symbol $\partial$ signals that other variables are held fixed. A value at a point is written $f_x(a, b)$ or $\frac{\partial f}{\partial x}(a, b)$ . The partial derivatives are themselves functions, defined at the points where the limits exist.

In plain words Several notations, one idea. The subscript or the letter under $\partial$ names the variable that moves.

Example For $f(x, y, z) = x e^{yz}$ , defined on all of $\mathbb{R}^3$ ,

\begin{gathered} \frac{\partial f}{\partial x} = e^{yz}, \qquad \frac{\partial f}{\partial y} = xz e^{yz}, \\[0.4em] \frac{\partial f}{\partial z} = xy e^{yz}. \end{gathered}

Domains matter too. For $g(x, y) = \ln(x - y)$ , defined on the open set $x > y$ , the partials $g_x = \frac{1}{x - y}$ and $g_y = -\frac{1}{x - y}$ hold only on that set.

Higher partial derivatives

Definition The partial derivatives of $f_x$ and $f_y$ , where they exist, are the second partial derivatives of $f$ . Writing the variables of differentiation in the order they are applied,

\begin{aligned} f_{xx} &= \frac{\partial^2 f}{\partial x^2}, & f_{xy} &= \frac{\partial^2 f}{\partial y \, \partial x}, \\[0.4em] f_{yx} &= \frac{\partial^2 f}{\partial x \, \partial y}, & f_{yy} &= \frac{\partial^2 f}{\partial y^2}. \end{aligned}

So $f_{xy} = (f_x)_y$ means differentiate first in $x$ , then in $y$ . In the $\partial$ notation the order reads from right to left. Third and higher partials are defined by repeating the process.

In plain words Take a partial derivative, then take a partial derivative of the result. Subscripts are read left to right in the order you differentiate.

Example Let $f(x, y) = x^3 + x^2 y^3 - 2y^2$ . Then

\begin{aligned} f_x &= 3x^2 + 2xy^3, \\[0.4em] f_y &= 3x^2 y^2 - 4y, \end{aligned}

and

\begin{aligned} f_{xx} &= 6x + 2y^3, & f_{xy} &= 6xy^2, \\[0.4em] f_{yx} &= 6xy^2, & f_{yy} &= 6x^2 y - 4. \end{aligned}

At $(2, 1)$ these give $f_{xx} = 14$ , $f_{yy} = 20$ and $f_{xy} = f_{yx} = 12$ .

Mixed partials

Definition The mixed partial derivatives of $f(x, y)$ are $f_{xy}$ and $f_{yx}$ . They come from differentiating in the two variables in opposite orders. Without extra hypotheses they need not be equal, even at a single point.

In plain words Changing the order of differentiation usually changes nothing, but not always, so equality has to be justified.

Example Let

\begin{gathered} f(x, y) = \frac{xy(x^2 - y^2)}{x^2 + y^2}, \\[0.4em] (x, y) \ne (0, 0), \qquad f(0, 0) = 0. \end{gathered}

Fix $y \ne 0$ . Since $f(0, y) = 0$ , the definition gives

\begin{aligned} f_x(0, y) &= \lim_{h \to 0} \frac{f(h, y)}{h} \\[0.4em] &= \lim_{h \to 0} \frac{y(h^2 - y^2)}{h^2 + y^2} = -y. \end{aligned}

The same formula holds at $y = 0$ , because $f$ is $0$ on the whole $x$ -axis and so $f_x(0, 0) = 0$ . A similar computation gives $f_y(x, 0) = x$ for every $x$ . Now $f_{xy}(0, 0)$ is the derivative of $f_x(0, y)$ in $y$ at $y = 0$ , and $f_{yx}(0, 0)$ is the derivative of $f_y(x, 0)$ in $x$ at $x = 0$ , so

f_{xy}(0, 0) = -1, \qquad f_{yx}(0, 0) = 1.

The mixed partials disagree at the origin.

Clairaut’s theorem

Definition Let $f$ be defined on an open disk $D$ containing the point $(a, b)$ . If the mixed partials $f_{xy}$ and $f_{yx}$ both exist at every point of $D$ and are continuous on $D$ , then

f_{xy}(a, b) = f_{yx}(a, b).

More generally, if all partial derivatives of $f$ of order up to $k$ are continuous on an open set, then on that set any partial derivative of order at most $k$ is the same whatever the order of differentiation, so for example $f_{xyy} = f_{yxy} = f_{yyx}$ when $k \ge 3$ .

In plain words When the mixed partials are continuous, order does not matter. The counterexample above escapes the theorem because its mixed partials are not continuous at the origin. Away from the origin its $f_{xy}$ equals $1$ on the $x$ -axis and $-1$ on the $y$ -axis, while $f_{xy}(0, 0) = -1$ .

Example Take $f(x, y) = \sin(xy)$ . Then $f_x = y\cos(xy)$ and $f_y = x\cos(xy)$ , so

f_{xy} = \cos(xy) - xy\sin(xy) = f_{yx}.

Both are continuous on all of $\mathbb{R}^2$ , exactly as Clairaut’s theorem predicts.

Partial differential equations

Definition A partial differential equation is an equation relating an unknown function of two or more variables to some of its partial derivatives. A solution on an open set is a function with the needed partial derivatives there that makes the equation true at every point of the set. Three classical examples, with constants $\alpha \ne 0$ and $c \ne 0$ , are

\begin{gathered} u_{xx} + u_{yy} = 0 \quad \text{(Laplace)}, \\[0.4em] u_t = \alpha^2 u_{xx} \quad \text{(heat)}, \\[0.4em] u_{tt} = c^2 u_{xx} \quad \text{(wave)}. \end{gathered}

In plain words The unknown is a whole function, and the equation constrains how its partial derivatives fit together. Checking a proposed solution is just computing partials and substituting.

Example The function $u(x, t) = e^{-t} \sin x$ solves the heat equation with $\alpha = 1$ on all of $\mathbb{R}^2$ . Indeed $u_t = -e^{-t}\sin x$ , and $u_x = e^{-t}\cos x$ gives $u_{xx} = -e^{-t}\sin x$ , the same function. It describes a sine-shaped temperature profile that fades over time.

Check yourself

Let $f(x, y) = xy^2 + 3x$ . Find the slope of the trace of its graph in the plane $x = 2$ at $y = 1$ , and the slope of the trace in the plane $y = 1$ at $x = 2$ .

Show the answer

The trace in the plane $x = 2$ is controlled by $f_y = 2xy$ , so its slope at $y = 1$ is $f_y(2, 1) = 4$ . The trace in the plane $y = 1$ is controlled by $f_x = y^2 + 3$ , so its slope at $x = 2$ is $f_x(2, 1) = 4$ .

Show that $u(x, y) = \ln(x^2 + y^2)$ satisfies Laplace’s equation on its domain.

Show the answer

The domain is $\mathbb{R}^2$ without the origin, an open set. There $u_x = \frac{2x}{x^2 + y^2}$ , and the quotient rule gives

\begin{aligned} u_{xx} &= \frac{2(y^2 - x^2)}{(x^2 + y^2)^2}, \\[0.4em] u_{yy} &= \frac{2(x^2 - y^2)}{(x^2 + y^2)^2}. \end{aligned}

Their sum is $0$ at every point of the domain.

Let $f(x, y) = \dfrac{xy}{x^2 + y^2}$ for $(x, y) \ne (0, 0)$ and $f(0, 0) = 0$ . Find $f_x(0, 0)$ and $f_y(0, 0)$ . Is $f$ continuous at the origin?

Show the answer

On the $x$ -axis $f(h, 0) = 0$ , so the difference quotient $\frac{f(h, 0) - f(0, 0)}{h}$ is $0$ and $f_x(0, 0) = 0$ . In the same way $f_y(0, 0) = 0$ . Yet $f$ is not continuous at the origin, because along $y = x$ its values are $\frac{1}{2}$ . Both partial derivatives exist and $f$ is still discontinuous.

Partial Derivatives

Key terms

Varying one coordinate

Slopes of traces

Partial derivative notation

Higher partial derivatives

Mixed partials

Clairaut’s theorem

Partial differential equations

Check yourself

How this lesson connects