Higher Partial Derivatives
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 4.3: “Partial Derivatives” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/4-3-partial-derivatives |
| Textbook used in class | Stewart, Calculus, Section 14.3: “Partial Derivatives” (Examples 6, 7) |
Quick Reference
For $f(x,y)$, the four second partial derivatives are:
$$f_{xx} = \frac{\partial^2 f}{\partial x^2}, \quad f_{xy} = \frac{\partial^2 f}{\partial y\,\partial x}, \quad f_{yx} = \frac{\partial^2 f}{\partial x\,\partial y}, \quad f_{yy} = \frac{\partial^2 f}{\partial y^2}.$$
Clairaut’s theorem: If $f$ is defined near $(a,b)$ and $f_{xy}$ and $f_{yx}$ are both continuous there, then $f_{xy}(a,b) = f_{yx}(a,b)$.
Laplace equation: $f_{xx} + f_{yy} = 0$ (or $\nabla^2 f = 0$). A function satisfying this is called harmonic.
Motivation
Second partial derivatives arise in physics and geometry in the same way that second derivatives do in one-variable calculus. The concavity of a surface in the $x$-direction is $f_{xx}$; in the $y$-direction, it is $f_{yy}$. Mixed partials $f_{xy}$ appear in the second derivative test for optimization and in partial differential equations describing heat flow, wave propagation, and electrostatics.
Clairaut’s theorem is the key theoretical fact: for the vast majority of functions (specifically, when the mixed partials are continuous), the order of differentiation does not matter.
Key Concepts
1. Computing Second Partial Derivatives
Differentiate the first partials just as you would differentiate any function:
- $f_{xx}$: differentiate $f_x$ with respect to $x$.
- $f_{yy}$: differentiate $f_y$ with respect to $y$.
- $f_{xy}$: differentiate $f_x$ with respect to $y$.
- $f_{yx}$: differentiate $f_y$ with respect to $x$.
2. Clairaut’s Theorem
For smooth functions (continuous second partials), $f_{xy} = f_{yx}$. In practice, this means you only need to compute one of the two mixed partials. The fact is non-trivial: there exist functions where $f_{xy}(0,0) \neq f_{yx}(0,0)$, but only when the mixed partials are discontinuous there.
3. Partial Differential Equations
A function satisfies a partial differential equation (PDE) if a relation involving its partial derivatives holds everywhere on a domain. The most important example in this course is the Laplace equation $f_{xx} + f_{yy} = 0$.
Worked Examples
Example 1. Find all four second partials of $f(x,y) = x^4 y^2 - x^2 y^5$.
First partials: $f_x = 4x^3 y^2 - 2xy^5$, $f_y = 2x^4 y - 5x^2 y^4$.
$$f_{xx} = 12x^2 y^2 - 2y^5, \quad f_{yy} = 2x^4 - 20x^2 y^3.$$ $$f_{xy} = 8x^3 y - 10xy^4, \quad f_{yx} = 8x^3 y - 10xy^4.$$
Clairaut’s theorem is confirmed: $f_{xy} = f_{yx}$.
Example 2. Show that $f(x,y) = e^x\sin y$ satisfies the Laplace equation $f_{xx}+f_{yy}=0$.
$f_x = e^x\sin y$, $f_{xx} = e^x\sin y$.
$f_y = e^x\cos y$, $f_{yy} = -e^x\sin y$.
$f_{xx}+f_{yy} = e^x\sin y - e^x\sin y = 0$. Confirmed.
$f_{xy}$ always equals $f_{yx}$, no matter what. Clairaut’s theorem requires the mixed partials to be continuous. There are pathological functions where $f_{xy}(0,0) \neq f_{yx}(0,0)$. However, for all functions built from elementary operations (polynomials, exponentials, trig), the mixed partials are continuous everywhere, so equality is guaranteed. In this course you may assume Clairaut’s theorem applies unless told otherwise.
Leveled Practice
Problem 1. Compute all second partial derivatives of $f(x,y) = x\ln(1+xy)$.
Show answer
$f_x = \ln(1+xy) + \dfrac{xy}{1+xy}$, $f_y = \dfrac{x^2}{1+xy}$.
$f_{xx}$: differentiate $f_x$ w.r.t. $x$: $\dfrac{y}{1+xy} + \dfrac{y(1+xy) - xy\cdot y}{(1+xy)^2} = \dfrac{2y}{1+xy} - \dfrac{y^2 x}{(1+xy)^2}$.
(This simplifies further but the computation is the key skill.)
$f_{yy}$: differentiate $f_y = x^2(1+xy)^{-1}$ w.r.t. $y$: $f_{yy} = -x^3(1+xy)^{-2}$.
$f_{xy} = f_{yx}$: differentiate $f_y = x^2/(1+xy)$ w.r.t. $x$: $\dfrac{2x(1+xy) - x^2 y}{(1+xy)^2} = \dfrac{2x + x^2 y}{(1+xy)^2}$.