Using Clairaut's Theorem to Reduce Computation

Why the Order of Differentiation Can Matter

For a function of two variables, a mixed second partial derivative is obtained by differentiating once with respect to each variable. In principle, there are two orders: differentiate first with respect to $x$ and then $y$, or first with respect to $y$ and then $x$. Each order may lead to a different intermediate expression and amount of work.

Clairaut’s theorem gives a way to reduce that work. When the mixed second partial derivatives are continuous in a neighborhood of a point, they are equal there. Thus, if the goal is to find the mixed partial at that point, it is enough to calculate whichever order is more convenient—provided the theorem’s conditions have been checked.

Key idea: Continuity of the mixed second partial derivatives near a point permits you to replace one order of mixed differentiation with the other. The equality is justified by the theorem’s hypotheses, not merely by a calculation that happens to produce matching answers.

Notation and the Theorem’s Conditions

For a function $f(x,y)$, the notation $f_{xy}$ means differentiate $f$ first with respect to $x$, then differentiate the result with respect to $y$. In other words, $f_{xy}=(f_x)_y$. Similarly, $f_{yx}=(f_y)_x$. The subscripts record the order: the rightmost differentiation is performed first.

One useful sufficient version of Clairaut’s theorem uses an open rectangle around the point. The rectangle ensures that the function and its relevant derivatives are considered on a two-dimensional neighborhood, not just at the point itself.

Theorem: Suppose $f_{xy}$ and $f_{yx}$ exist and are continuous on an open rectangle containing $(a,b)$. Then $$ f_{xy}(a,b)=f_{yx}(a,b). $$ The same conclusion holds at every point of that rectangle. These are sufficient conditions; the theorem does not say that continuity is necessary in every case where the mixed partials happen to be equal.

For many familiar functions—such as polynomials, exponential and trigonometric functions, and logarithms on their domains—the relevant second partial derivatives are continuous wherever the expressions are defined. In those cases, the main task is to check that the point lies in the domain and that the derivatives are continuous on some open rectangle around it.

Why One Calculation Is Enough

The reason the method saves work is direct. Under the stated continuity conditions, the theorem says that the two mixed partials agree throughout a neighborhood of $(a,b)$. In particular, they agree at the point where we want their value. Therefore, after finding one of them, its value at $(a,b)$ is already the value of the other.

f_{xy}=f_{yx}\text{ on a neighborhood of }(a,b) \quad\Longrightarrow\quad f_{xy}(a,b)=f_{yx}(a,b).

This reasoning depends on the theorem’s assumptions. A calculation of $f_{xy}(a,b)$, by itself, says nothing about $f_{yx}(a,b)$. The inference becomes valid only after establishing that the mixed partials exist and satisfy the required continuity condition near the point.

Worked Examples

Worked Example: A Polynomial-Exponential-Trigonometric Sum

Let $f(x,y)=x^2y^3+e^x\sin y$. Find $f_{yx}(1,0)$ by computing only the order $f_{xy}$.

This function is a sum of products of polynomials, an exponential, and a trigonometric function. Its second partial derivatives are continuous on $\mathbb R^2$, so Clairaut’s theorem applies near $(1,0)$.

First differentiate with respect to $x$:

f_x(x,y)=2xy^3+e^x\sin y.

Now differentiate this result with respect to $y$:

f_{xy}(x,y) =2x(3y^2)+e^x\cos y =6xy^2+e^x\cos y.

Evaluate at $(1,0)$. Since $0^2=0$, $e^1=e$, and $\cos 0=1$,

f_{xy}(1,0)=6(1)(0)^2+e^1\cos 0=0+e=e.

The theorem’s conditions hold near the point, so $f_{yx}(1,0)=f_{xy}(1,0)=e$. The requested value is therefore $e$, without separately computing $f_y$ and then differentiating it with respect to $x$.

Worked Example: An Exponential with a Product in Its Exponent

Let $g(x,y)=e^{xy}+x^2y^3$. Find $g_{xy}(0,1)$ by computing $g_{yx}$.

Both terms have continuous second partial derivatives on $\mathbb R^2$. It is therefore valid to compute the order that the question asks us to use and infer the other order.

Differentiate $g$ first with respect to $y$. By the chain rule, the derivative of $e^{xy}$ with respect to $y$ is $xe^{xy}$:

g_y(x,y)=xe^{xy}+3x^2y^2.

Now differentiate with respect to $x$. The product rule gives $\frac{\partial}{\partial x}(xe^{xy})=e^{xy}+x(ye^{xy})$, while $\frac{\partial}{\partial x}(3x^2y^2)=6xy^2$. Hence

g_{yx}(x,y)=e^{xy}+xye^{xy}+6xy^2.

Substitute $x=0$ and $y=1$ into each term:

g_{yx}(0,1) =e^{0\cdot 1}+(0)(1)e^{0\cdot 1}+6(0)(1)^2 =1+0+0 =1.

Because the mixed partials are continuous near $(0,1)$, Clairaut’s theorem gives $g_{xy}(0,1)=g_{yx}(0,1)=1$. Thus the requested value is $1$.

Worked Example: A Logarithm on Its Domain

Let $h(x,y)=\ln(1+x^2+y^2)$. Find $h_{yx}(1,0)$ by computing $h_{xy}$.

The logarithm’s argument satisfies $1+x^2+y^2\geq 1>0$ for every real $x,y$. Thus $h$ is defined on all of $\mathbb R^2$, and its second partial derivatives are continuous there. The theorem applies at $(1,0)$.

Write $D=1+x^2+y^2$ to keep the denominator visible. Differentiating first with respect to $x$ gives

h_x(x,y)=\frac{2x}{D} =\frac{2x}{1+x^2+y^2}.

Now differentiate with respect to $y$. Here $2x$ is constant with respect to $y$, and the derivative of $D^{-1}$ is $-D^{-2}(2y)$. Therefore

h_{xy}(x,y) =2x\bigl(-D^{-2}(2y)\bigr) =-\frac{4xy}{(1+x^2+y^2)^2}.

At $(1,0)$, the numerator is $-4(1)(0)=0$, while the denominator is $(1+1^2+0^2)^2=4$. Thus

h_{xy}(1,0)=\frac{0}{4}=0.

Clairaut’s theorem now gives $h_{yx}(1,0)=h_{xy}(1,0)=0$. The value is checked by substitution into the calculated expression and by confirming that the continuity condition applies at the point.

A Reliable Procedure

For a mixed-partial evaluation, separate the justification from the differentiation. First establish that the theorem applies near the point. Then choose one order, carry out that differentiation carefully, and evaluate the resulting expression. Finally, state explicitly how the theorem transfers the value to the other order.

Check the neighborhood.
Confirm that the function is defined on an open rectangle around the point and that both mixed second partial derivatives exist and are continuous there.

Choose an order.
Compute the mixed partial that appears simpler or is requested directly. Remember that $f_{xy}$ means $x$ first and then $y$.

Evaluate at the point.
Substitute both coordinates into the expression you derived, showing the intermediate values.

Use the theorem to conclude.
With the continuity conditions established, identify the value of the other mixed partial at the same point.

Common Mistake: Assuming Equality Without Checking Conditions

A tempting but incomplete argument is: “I computed one mixed partial, so the other one has the same value.” That conclusion does not follow from the calculation alone. The continuity hypotheses are what justify replacing one order with the other.

For example, suppose a calculation produces $f_{xy}(a,b)=7$. Writing $f_{yx}(a,b)=7$ immediately is an unjustified step unless the required conditions for Clairaut’s theorem have been verified. The corrected reasoning is to establish that both mixed partials exist and are continuous on a neighborhood of $(a,b)$, invoke the theorem, and then conclude $f_{yx}(a,b)=f_{xy}(a,b)=7$.

A related error is to check continuity only at the point without confirming that the theorem version being used allows that weaker condition. The sufficient version stated here asks for continuity on an open rectangle around the point. Using a clear, stated set of hypotheses prevents accidentally applying a theorem under conditions that have not been checked.

Why This Reduction Is Useful

Mixed-partial calculations can involve product rules, chain rules, and several terms. Taking a derivative in one order may create an expression that is more cumbersome to differentiate again than the alternative. Clairaut’s theorem allows a choice: once continuity near the point is assured, compute the order that keeps the algebra manageable.

The theorem also clarifies what a computation can and cannot establish. A single order gives a value for that order. The continuity condition supplies the logical bridge to the other order. Keeping these two steps distinct makes the work shorter without making the reasoning less rigorous.

Check Your Understanding

For each question, pay attention both to the order of differentiation and to the hypotheses that permit an inference.

In the notation $f_{yx}$, which variable is differentiated first?
State the continuity condition in the version of Clairaut’s theorem used here.
If $f_{xy}(a,b)=4$ and the theorem’s hypotheses hold near $(a,b)$, what is $f_{yx}(a,b)$?
Why is computing $f_{xy}(a,b)$ not, by itself, enough to conclude the value of $f_{yx}(a,b)$?
For $q(x,y)=e^{x+y}$, which fact lets you use Clairaut’s theorem near any point in $\mathbb R^2$?