Higher Partial Derivatives

Section 4.2 Higher Partial Derivatives

The partial derivatives of the function \(z=f(x,y)\) are themselves functions of two variables. Thus they can be differentiated further, giving the second partial derivatives, the third partial derivatives etc. Common notations for the second partial derivatives include:

\begin{alignat*}{2} \dfrac{\partial}{\partial x} \left( \dfrac{\partial f}{\partial x} \right) \amp = \dfrac{\partial^2 f}{\partial x^2} \amp = (f_x)_x \amp = f_{xx}\\ \dfrac{\partial}{\partial y} \left( \dfrac{\partial f}{\partial x} \right) \amp = \dfrac{\partial^2 f}{\partial y \partial x} \amp = (f_x)_y \amp = f_{xy}\\ \dfrac{\partial}{\partial x} \left( \dfrac{\partial f}{\partial y} \right) \amp = \dfrac{\partial^2 f}{\partial x \partial y} \amp = (f_y)_x \amp = f_{yx}\\ \dfrac{\partial}{\partial y} \left( \dfrac{\partial f}{\partial y} \right) \amp = \dfrac{\partial^2 f}{\partial y^2} \amp = (f_y)_y \amp = f_{yy} \end{alignat*}

Example 4.2.1.

Find the second partial derivatives for the function

\begin{equation*} f(x,y) = \sin(x) \cos(y)\text{.} \end{equation*}

Answer.

\(f_{xx}(x,y) = -\sin(x) \cos(y)\text{,}\) \(f_{xy}(x,y) = -\cos(x) \sin(y)\text{,}\) \(f_{yx}(x,y) = -\cos(x) \sin(y)\) and \(f_{yy}(x,y) = -\sin(x) \cos(y).\)

Solution.

Begin by finding the first partial derivatives. Here

\begin{align*} \amp f_x(x,y) = \cos(x)\cos(y) \amp \text{and} \amp \amp f_y(x,y) = - \sin(x) \sin(y)\text{.} \end{align*}

Now differentiate \(f_x(x,y)\) firstly with respect to \(x\) to find \(f_{xx}(x,y)\) and then with respect to \(y\) to find \(f_{xy}(x,y)\text{.}\) Thus,

\begin{align*} f_{xx}(x,y) \amp = -\sin(x) \cos(y), \, \, \text{and}\\ f_{xy}(x,y) \amp = -\cos(x) \sin(y). \end{align*}

Next differentiate \(f_y(x,y)\) with respect to \(x\) to find \(f_{yx}(x,y)\) and then with respect to \(y\) to find \(f_{yy}(x,y)\text{.}\) Thus,

\begin{align*} f_{yx}(x,y) \amp = -\cos(x) \sin(y), \, \, \text{and}\\ f_{yy}(x,y) \amp = -\sin(x) \cos(y). \end{align*}

Notice that for this function \(f_{xy}(x,y) = f_{yx}(x,y)\text{.}\)

Example 4.2.2.

Calculate \(g_{xxy}\text{,}\) \(g_{xyx}\) and \(g_{yxx}\) when \(g(x)=xy^2 + \dfrac{y}{x^2}\text{.}\)

Answer.

\(g_{xxy} = 6x^{-4}\text{,}\) \(g_{xyx} = 6x^{-4}\) and \(g_{yxx} = 6x^{-4}\text{.}\)

Solution.

Begin by writing the function in the form

\begin{equation*} g(x,y) = xy^2 + yx^{-2}\text{.} \end{equation*}

Then the first partial derivatives are

\begin{align*} \amp g_x = y^2 - 2yx^{-3} \amp \text{and} \amp \amp g_y = 2xy+x^{-2} \end{align*}

and hence the second partial derivatives are

\begin{equation*} g_{xx}=6yx^{-4}, \; g_{xy} = 2y-2x^{-3}, \; g_{yx} = 2y-2x^{-3}, \; g_{yy} = 2x\text{.} \end{equation*}

Differentiating \(g_{xx}\) with respect to \(y\text{,}\) \(g_{xy}\) with respect to \(x\) and \(g_{yx}\) with respect to \(x\) gives

\begin{equation*} g_{xxy} = 6x^{-4}, \; g_{xyx} = 6x^{-4}, \; g_{yxx} = 6x^{-4}\text{.} \end{equation*}

The above instances have provided examples of the following general result.

Theorem 4.2.3. Clairaut's Theorem.

If for the function \(f(x,y)\) both \(f_{xy}\) and \(f_{yx}\) are continuous on some domain \(D\text{,}\) then on that domain

\begin{equation*} f_{xy}(x,y) = f_{yx}(x,y)\text{.} \end{equation*}

Clairaut's Theorem can be extended to higher partial derivatives and to functions of more than two variables.

Example 4.2.4.

Calculate all first and second order partial derivatives for the function

\begin{equation*} g(x,y,z) = \dfrac{x^2 + 3y^2}{1+2z}\text{.} \end{equation*}

Solution.

Even though \(g\) is a function of \(3\) variables, Clairaut's Theorem still holds. Thus there will be only 6 distinct second partial derivatives, i.e. \(g_{xx}\text{,}\) \(g_{xy}\text{,}\) \(g_{xz}\text{,}\) \(g_{yy}\text{,}\) \(g_{yz}\text{,}\) \(g_{zz}\text{.}\)

Now

\begin{equation*} g_x = \dfrac{2x}{1+2z}, \,\, g_y = \dfrac{6y}{1+2z}, \,\, g_z = \dfrac{-2(x^2+3y^2)}{(1+2z)^2} \end{equation*}

and so

\begin{equation*} g_{xx} = \dfrac{2}{1+2z}, \,\, g_{xy} = 0, \,\, g_{xz} = \dfrac{-4x}{(1+2z)^2}, \end{equation*}

\begin{equation*} g_{yz} = \dfrac{-12y}{(1+2z)^2}, \,\, g_{yy} = \dfrac{6}{1+2z}, \,\, g_{zz} = \dfrac{8(x^2+3y^2)}{(1+2z)^3}. \end{equation*}

Exercises Example Tasks

1.

Find the second partial derivatives for the function \(z=xye^y\text{.}\)

2.

Calculate \(g_{xx},\,\) \(g_{xy}\) and \(g_{xyy}\) for \(g(x,y) = \dfrac{y}{1+x^2}\text{.}\)

3.

Let \(u(x,t) = e^{-t} \sin \left(\dfrac{x}{c}\right)\) where \(c\) is a constant and \(c>0\text{.}\) Determine if \(u\) satisfies

\begin{equation*} \dfrac{\partial u}{\partial t} = c^2 \dfrac{\partial^2 u}{\partial x^2} \end{equation*}

Remark 4.2.5. A little remark.

Computer algebra systems can find partial derivatives. For example, here are some examples of a queries to Wolfram Alpha that will work.