2.2 Partial Derivatives

Now that we have an idea of what functions of several variables are, and what a limit of such a function is, we can start to develop an idea of a derivative of a function of two or more variables. We will start with the notion of a partial derivative.

Definition 2.3

Let $f(x,y)$ be a real-valued function with domain $D$ in $\Real{2}$, and let $(a,b)$ be a point in $D$. Then the partial derivative of $\bm{f}$ at $\bm{(a,b)}$ with respect to $\bm{x}$, denoted by $\dfrac{\partial f}{\partial x} (a,b)$, is defined as

\begin{equation} \dfrac{\partial f}{\partial x} (a,b) = \lim_{h \to 0} \dfrac{f(a+h,b) - f(a,b)}{h} \tag{2.2}\end{equation}

and the partial derivative of $\bm{f}$ at $\bm{(a,b)}$ with respect to $\bm{y}$, denoted by $\dfrac{\partial f}{\partial y} (a,b)$, is defined as

\begin{equation} \dfrac{\partial f}{\partial y} (a,b) = \lim_{h \to 0} \dfrac{f(a,b+h) - f(a,b)}{h} ~. \tag{2.3}\end{equation}

Note: The symbol $\partial$ is pronounced “del”.^[1]

Recall that the derivative of a function $f(x)$ can be interpreted as the rate of change of that function in the (positive) $x$ direction. From the definitions above, we can see that the partial derivative of a function $f(x,y)$ with respect to $x$ or $y$ is the rate of change of $f(x,y)$ in the (positive) $x$ or $y$ direction, respectively. What this means is that the partial derivative of a function $f(x,y)$ with respect to $x$ can be calculated by treating the $y$ variable as a constant, and then simply differentiating $f(x,y)$ as if it were a function of $x$ alone, using the usual rules from single-variable calculus. Likewise, the partial derivative of $f(x,y)$ with respect to $y$ is obtained by treating the $x$ variable as a constant and then differentiating $f(x,y)$ as if it were a function of $y$ alone.

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Example 2.10

Find $\dfrac{\partial f}{\partial x} (x,y)$ and $\dfrac{\partial f}{\partial y} (x,y)$ for the function $f(x,y) = x^{2}y + y^3$.

Solution: Treating $y$ as a constant and differentiating $f(x,y)$ with respect to $x$ gives

\[\dfrac{\partial f}{\partial x} (x,y) = 2xy\]

and treating $x$ as a constant and differentiating $f(x,y)$ with respect to $y$ gives

\[\dfrac{\partial f}{\partial y} (x,y) = x^2 + 3y^2 ~.\]

width height 0.5pt We will often simply write $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ instead of $\dfrac{\partial f}{\partial x} (x,y)$ and $\dfrac{\partial f}{\partial y} (x,y)$.

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Example 2.11

Find $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ for the function $f(x,y) = \dfrac{\sin (xy^2)}{x^2 + 1}$.

Solution: Treating $y$ as a constant and differentiating $f(x,y)$ with respect to $x$ gives

\[\dfrac{\partial f}{\partial x} = \dfrac{(x^2 + 1)(y^2 \cos (xy^2)) - (2x)\,\sin (xy^2)}{(x^2 + 1)^2}\]

and treating $x$ as a constant and differentiating $f(x,y)$ with respect to $y$ gives

\[\dfrac{\partial f}{\partial y} = \dfrac{2xy\,\cos (xy^2)}{x^2 + 1} ~.\]

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Since both $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ are themselves functions of $x$ and $y$, we can take their partial derivatives with respect to $x$ and $y$. This yields the higher-order partial derivatives:

\begin{gather*} \dfrac{\partial^2 f}{\partial x^2} = \dfrac{\partial}{\partial x} \biggl( \dfrac{\partial f}{\partial x} \biggr) \qquad\dfrac{\partial^2 f}{\partial y^2} =\dfrac{\partial}{\partial y} \biggl( \dfrac{\partial f}{\partial y} \biggr)\\ \dfrac{\partial^2 f}{\partial y \, \partial x} = \dfrac{\partial}{\partial y} \biggl( \dfrac{\partial f}{\partial x} \biggr)\qquad \dfrac{\partial^2 f}{\partial x \, \partial y} = \dfrac{\partial}{\partial x} \biggl( \dfrac{\partial f}{\partial y} \biggr)\\ \dfrac{\partial^3 f}{\partial x^3} = \dfrac{\partial}{\partial x} \biggl( \dfrac{\partial^2 f}{\partial x^2} \biggr) \qquad\dfrac{\partial^3 f}{\partial y^3} =\dfrac{\partial}{\partial y} \biggl( \dfrac{\partial^2 f}{\partial y^2} \biggr)\\ \dfrac{\partial^3 f}{\partial y \, \partial x^2} = \dfrac{\partial}{\partial y} \biggl( \dfrac{\partial^2 f}{\partial x^2} \biggr)\qquad \dfrac{\partial^3 f}{\partial x \, \partial y^2} = \dfrac{\partial}{\partial x} \biggl( \dfrac{\partial^2 f}{\partial y^2} \biggr)\\ \dfrac{\partial^3 f}{\partial y^2 \, \partial x} = \dfrac{\partial}{\partial y} \biggl( \dfrac{\partial^2 f}{\partial y \, \partial x} \biggr)\qquad \dfrac{\partial^3 f}{\partial x^2 \, \partial y} = \dfrac{\partial}{\partial x} \biggl( \dfrac{\partial^2 f}{\partial x \, \partial y} \biggr)\\ \dfrac{\partial^3 f}{\partial x \, \partial y \, \partial x} = \dfrac{\partial}{\partial x} \biggl( \dfrac{\partial^2 f}{\partial y \, \partial x} \biggr)\qquad \dfrac{\partial^3 f}{\partial y \, \partial x \, \partial y} = \dfrac{\partial}{\partial y} \biggl( \dfrac{\partial^2 f}{\partial x \, \partial y} \biggr)\\ \vdots \end{gather*}

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Example 2.12

Find the partial derivatives $\dfrac{\partial f}{\partial x}$, $\dfrac{\partial f}{\partial y}$, $\dfrac{\partial^2 f}{\partial x^2}$, $\dfrac{\partial^2 f}{\partial y^2}$, $\dfrac{\partial^2 f}{\partial y \, \partial x}$ and $\dfrac{\partial^2 f}{\partial x \, \partial y}$ for the function $f(x,y) = e^{x^2 y} + xy^3$.

Solution: Proceeding as before, we have

\begin{alignat*}{2} \dfrac{\partial f}{\partial x} &= 2xy e^{x^2 y} + y^3\qquad\qquad & \dfrac{\partial f}{\partial y} &= x^2 e^{x^2 y} + 3xy^2\\ \dfrac{\partial^2 f}{\partial x^2} &= \dfrac{\partial}{\partial x} (2xy e^{x^2 y} + y^3)\qquad\qquad & \dfrac{\partial^2 f}{\partial y^2} &= \dfrac{\partial}{\partial y} (x^2 e^{x^2 y} + 3xy^2)\\ \phantom{\dfrac{\partial^2 f}{\partial x^2}} &= 2y e^{x^2 y} + 4x^2 y^2 e^{x^2 y}\qquad\qquad & \phantom{\dfrac{\partial^2 f}{\partial y^2}} &= x^4 e^{x^2 y} + 6xy\\ \dfrac{\partial^2 f}{\partial y \, \partial x} &= \dfrac{\partial}{\partial y} (2xy e^{x^2 y} + y^3)\qquad\qquad & \dfrac{\partial^2 f}{\partial x \, \partial y} &= \dfrac{\partial}{\partial x} (x^2 e^{x^2 y} + 3xy^2)\\ \phantom{\dfrac{\partial^2 f}{\partial y \, \partial x}} &= 2x e^{x^2 y} + 2x^3 y e^{x^2 y} + 3y^2\qquad\qquad & \phantom{\dfrac{\partial^2 f}{\partial x \, \partial y}} &= 2x e^{x^2 y} + 2x^3 y e^{x^2 y} + 3y^2 \end{alignat*}

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Higher-order partial derivatives that are taken with respect to different variables, such as $\frac{\partial^2 f}{\partial y \, \partial x}$ and $\frac{\partial^2 f}{\partial x \, \partial y}$, are called mixed partial derivatives. Notice in the above example that $\frac{\partial^2 f}{\partial y \, \partial x} = \frac{\partial^2 f}{\partial x \, \partial y}$. It turns that this will usually be the case. Specifically, whenever both $\frac{\partial^2 f}{\partial y \, \partial x}$ and $\frac{\partial^2 f}{\partial x \, \partial y}$ are continuous at a point $(a,b)$, then they are equal at that point.^[2] All the functions we will deal with will have continuous partial derivatives of all orders, so you can assume in the remainder of the text that

\[\frac{\partial^2 f}{\partial y \, \partial x} = \frac{\partial^2 f}{\partial x \, \partial y} ~~ \text{for all $(x,y)$ in the domain of $f$.}\]

In other words, it doesn’t matter in which order you take partial derivatives. This applies even to mixed partial derivatives of order $3$ or higher.

The notation for partial derivatives varies. All of the following are equivalent:

\begin{gather*} \dfrac{\partial f}{\partial x}:~ \ssub{f}{x}(x,y) ~, \quad \ssub{f}{1}(x,y) ~, \quad \ssub{D}{x}(x,y) ~, \quad \ssub{D}{1}(x,y)\\ \dfrac{\partial f}{\partial y}:~ \ssub{f}{y}(x,y) ~, \quad \ssub{f}{2}(x,y) ~, \quad \ssub{D}{y}(x,y) ~, \quad \ssub{D}{2}(x,y)\\ \dfrac{\partial^2 f}{\partial x^2}:~ \ssub{f}{xx}(x,y) ~, \quad \ssub{f}{11}(x,y) ~, \quad \ssub{D}{xx}(x,y) ~, \quad \ssub{D}{11}(x,y)\\ \dfrac{\partial^2 f}{\partial y^2}:~ \ssub{f}{yy}(x,y) ~, \quad \ssub{f}{22}(x,y) ~, \quad \ssub{D}{yy}(x,y) ~, \quad \ssub{D}{22}(x,y)\\ \dfrac{\partial^2 f}{\partial y \, \partial x}: ~ \ssub{f}{xy}(x,y) ~, \quad \ssub{f}{12}(x,y) ~, \quad \ssub{D}{xy}(x,y) ~, \quad \ssub{D}{12}(x,y)\\ \dfrac{\partial^2 f}{\partial x \, \partial y}: ~ \ssub{f}{yx}(x,y) ~, \quad \ssub{f}{21}(x,y) ~, \quad \ssub{D}{yx}(x,y) ~, \quad \ssub{D}{21}(x,y) \end{gather*}

Exercises

For Exercises 1-16, find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$.

$f(x,y) = x^2 + y^2$
$f(x,y) = \cos(x+y)$

2
$f(x,y) = \sqrt{x^2 + y + 4}$
$f(x,y) = \dfrac{x + 1}{y + 1}$

2
$f(x,y) = e^{xy} + xy$
$f(x,y) = x^2 - y^2 + 6xy + 4x - 8y + 2$

2
$f(x,y) = x^4$
$f(x,y) = x + 2y$

2
$f(x,y) = \sqrt{x^2 + y^2}$
$f(x,y) = \sin(x+y)$

2
$f(x,y) = \sqrt[3]{x^2 + y + 4}$
$f(x,y) = \dfrac{xy + 1}{x + y}$

2
$f(x,y) = e^{-(x^2 + y^2)}$
$f(x,y) = \ln (xy)$

2
$f(x,y) = \sin(xy)$
$f(x,y) = \tan(x+y)$

For Exercises 17-26, find $\frac{\partial^2 f}{\partial x^2}$, $\frac{\partial^2 f}{\partial y^2}$ and $\frac{\partial^2 f}{\partial y \, \partial x}$ (use Exercises 1-8, 14, 15).

$f(x,y) = x^2 + y^2$
$f(x,y) = \cos(x+y)$

2
$f(x,y) = \sqrt{x^2 + y + 4}$
$f(x,y) = \dfrac{x + 1}{y + 1}$

2
$f(x,y) = e^{xy} + xy$
$f(x,y) = x^2 - y^2 + 6xy + 4x - 8y + 2$

2
$f(x,y) = x^4$
$f(x,y) = x + 2y$

2
$f(x,y) = \ln (xy)$
$f(x,y) = \sin(xy)$

Show that the function $f(x,y)=\sin(x+y) + \cos(x-y)$ satisfies the wave equation

\[\frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{\partial y^2} = 0 ~.\]

The wave equation is an example of a partial differential equation.
Let $u$ and $v$ be twice-differentiable functions of a single variable, and let $c\ne 0$ be a constant. Show that $f(x,y)=u(x+cy)+v(x-cy)$ is a solution of the general one-dimensional wave equation^[3]

\[\frac{\partial^2 f}{\partial x^2} - \frac{1}{c^2}\,\frac{\partial^2 f}{\partial y^2} = 0 ~.\]

It is not a Greek letter. The symbol was first used by the mathematicians A. Clairaut and L. Euler around 1740, to distinguish it from the letter $d$ used for the “usual” derivative. ↩
See pp. 214-216 in tm for a proof. ↩
Conversely, it turns out that any solution must be of this form. See Ch. 1 in wei. ↩