Clairaut's Theorem on Higher Order Partial Derivatives Examples 1

# Clairaut's Theorem on Higher Order Partial Derivatives

Recall from the Clairaut's Theorem on Higher Order Partial Derivatives page that Clairaut's theorem says that if $z = f(x, y)$ is a two variable real-valued function defined on a disk $\mathcal D$ containing the point $(a, b)$ then if the second order mixed partial derivatives of $f$ are continuous on $\mathcal D$ then:

(1)
\begin{align} \quad \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial x \partial y} \end{align}

We will now look at some examples regarding Clairaut's theorem.

## Example 1

Consider the function $z = x^2 y^4 - 2xy^4 + 2(xy + 1)$. Verify that the second order mixed partial derivatives of this function are equal.**

We first compute the first partial derivatives of this function to get that:

(2)
\begin{align} \quad \frac{\partial z}{\partial x} = 2xy^4 - 2y^4 + 2y \quad , \quad \frac{\partial z}{\partial y} = 4x^2 y^3 - 8 xy^3 + 2x \end{align}

Now we compute the second order partial derivatives of this function:

(3)
\begin{align} \quad \frac{\partial^2 z}{\partial y \partial x} = 8xy^3 - 8y^3 + 2 \end{align}
(4)
\begin{align} \quad \frac{\partial^2 z}{\partial x \partial y} = 8x y^3 - 8y^3 + 2 \end{align}

Therefore we see that $\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial x \partial y}$. This should not be surprising since the given function is merely a polynomial and polynomials are continuous everywhere.