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CLP-3 Multivariable Calculus

Joel Feldman, Andrew Rechnitzer, Elyse Yeager

Section 2.3 Higher Order Derivatives

You have already observed, in your first Calculus course, that if \(f(x)\) is a function of \(x\text{,}\) then its derivative, \(\diff{f}{x}(x)\text{,}\) is also a function of \(x\text{,}\) and can be differentiated to give the second order derivative \(\difftwo{f}{x}(x)\text{,}\) which can in turn be differentiated yet again to give the third order derivative, \(f^{(3)}(x)\text{,}\) and so on.
We can do the same for functions of more than one variable. If \(f(x,y)\) is a function of \(x\) and \(y\text{,}\) then both of its partial derivatives, \(\pdiff{f}{x}(x,y)\) and \(\pdiff{f}{y}(x,y)\) are also functions of \(x\) and \(y\text{.}\) They can both be differentiated with respect to \(x\) and they can both be differentiated with respect to \(y\text{.}\) So there are four possible second order derivatives. Here they are, together with various alternate notations.
\begin{alignat*}{2} \pdiff{}{x}\left(\pdiff{f}{x}\right)(x,y) &=\frac{\partial^2 f}{\partial x^2}(x,y) &= f_{xx}(x,y)\\ \pdiff{}{y}\left(\pdiff{f}{x}\right)(x,y) &=\frac{\partial^2\ f}{\partial y\partial x}(x,y) &= f_{xy}(x,y)\\ \pdiff{}{x}\left(\pdiff{f}{y}\right)(x,y) &=\frac{\partial^2\ f}{\partial x\partial y}(x,y) &= f_{yx}(x,y)\\ \pdiff{}{y}\left(\pdiff{f}{y}\right)(x,y) &=\frac{\partial^2 f}{\partial y^2}(x,y) &= f_{yy}(x,y) \end{alignat*}
In \(\frac{\partial^2\ f}{\partial y\,\partial x} =\frac{\partial^2}{\partial y\,\partial x}f\text{,}\) the derivative closest to \(f\text{,}\) in this case \(\pdiff{}{x}\text{,}\) is applied first.
In \(f_{xy}\text{,}\) the derivative with respect to the variable closest to \(f\text{,}\) in this case \(x\text{,}\) is applied first.

Example 2.3.1.

Let \(f(x,y) = e^{my}\cos(nx)\text{.}\) Then
\begin{align*} f_x &= -n e^{my}\sin(nx) & f_y &= m e^{my}\cos(nx)\\ f_{xx} &= -n^2 e^{my}\cos(nx) & f_{yx} &= -m n e^{my}\sin(nx)\\ f_{xy} &= -m n e^{my}\sin(nx) & f_{yy} &= m^2 e^{my}\cos(nx) \end{align*}

Example 2.3.2.

Let \(f(x,y) = e^{\al x+\be y}\text{.}\) Then
\begin{align*} f_x &= \al e^{\al x+\be y} & f_y &= \be e^{\al x+\be y}\\ f_{xx} &= \al^2 e^{\al x+\be y} & f_{yx} &= \be \al e^{\al x+\be y}\\ f_{xy} &= \al \be e^{\al x+\be y} & f_{yy} &= \be^2 e^{\al x+\be y} \end{align*}
More generally, for any integers \(m,n\ge 0\text{,}\)
\begin{equation*} \frac{\partial^{m+n} f}{\partial x^m\, \partial y^n} = \al^m\be^n e^{\al x+\be y} \end{equation*}

Example 2.3.3.

If \(f(x_1,x_2,x_3,x_4) = x_1^4\, x_2^3\, x_3^2\, x_4\text{,}\) then
\begin{align*} \frac{\partial^4\ f} {\partial x_1\, \partial x_2\,\partial x_3\,\partial x_4} &= \frac{\partial^3 \ } {\partial x_1\, \partial x_2\,\partial x_3} \left( x_1^4\, x_2^3\, x_3^2\right)\\ &= \frac{\partial^2 \ } {\partial x_1\, \partial x_2} \left( 2\ x_1^4\, x_2^3\, x_3\right)\\ &= \pdiff{}{x_1} \left( 6\ x_1^4\, x_2^2\, x_3\right)\\ &= 24\ x_1^3\, x_2^2\, x_3 \end{align*}
and
\begin{align*} \frac{\partial^4\ f} {\partial x_4\, \partial x_3\,\partial x_2\,\partial x_1} &= \frac{\partial^3 \ } {\partial x_4\, \partial x_3\,\partial x_2} \left( 4 x_1^3\, x_2^3\, x_3^2\,x_4\right)\\ &= \frac{\partial^2 \ } {\partial x_4\, \partial x_3} \left( 12\ x_1^3\, x_2^2\, x_3^2\,x_4\right)\\ &= \pdiff{}{x_4} \left( 24\ x_1^3\, x_2^2\, x_3\,x_4\right)\\ &= 24\ x_1^3\, x_2^2\, x_3 \end{align*}
Notice that in Example 2.3.1,
\begin{equation*} f_{xy}= f_{yx} = -m n e^{my}\sin(nx) \end{equation*}
and in Example 2.3.2
\begin{equation*} f_{xy}= f_{yx} = \al \be e^{\al x+\be y} \end{equation*}
and in Example 2.3.3
\begin{equation*} \frac{\partial^4\ f} {\partial x_1\, \partial x_2\,\partial x_3\,\partial x_4} = \frac{\partial^4\ f} {\partial x_4\, \partial x_3\,\partial x_2\,\partial x_1} = 24\ x_1^3\, x_2^2\, x_3 \end{equation*}
In all of these examples, it didn’t matter what order we took the derivatives in. The following theorem
 1 
The history of this important theorem is pretty convoluted. See “A note on the history of mixed partial derivatives” by Thomas James Higgins which was published in Scripta Mathematica 7 (1940), 59-62. The Theorem is named for Alexis Clairaut (1713--1765), a French mathematician, astronomer, and geophysicist, and Hermann Schwarz (1843--1921), a German mathematician.
shows that this was no accident.

Subsection 2.3.1 Optional — The Proof of Theorem 2.3.4

Subsubsection 2.3.1.1 Outline

Here is an outline of the proof of Theorem 2.3.4. The (numbered) details are in the subsection below. Fix real numbers \(x_0\) and \(y_0\) and define
\begin{equation*} F(h,k) =\frac{1}{hk}\big[f(x_0+h,y_0+k)-f(x_0,y_0+k)-f(x_0+h,y_0)+f(x_0,y_0)\big] \end{equation*}
We define \(F(h,k)\) in this way because both partial derivatives \(\frac{\partial^2 f}{\partial x\partial y}(x_0,y_0)\) and \(\frac{\partial^2 f}{\partial y\partial x}(x_0,y_0)\) are limits of \(F(h,k)\) as \(h,k\rightarrow 0\text{.}\) Precisely, we show in item (1) in the details below that
\begin{align*} \pdiff{}{y} \pdiff{f}{x}(x_0,y_0) &= \lim_{k\rightarrow 0}\lim_{h\rightarrow 0}F(h,k)\\ \pdiff{}{x} \pdiff{f}{y}(x_0,y_0) &= \lim_{h\rightarrow 0}\lim_{k\rightarrow 0}F(h,k) \end{align*}
Note that the two right hand sides here are identical except for the order in which the limits are taken.
Now, by the mean value theorem (four times),
\begin{align*} F(h,k)\ &\eqf{(2)}\ \frac{1}{h} \left[\pdiff{f}{y}(x_0+h,y_0+\theta_1k) -\pdiff{f}{y}(x_0,y_0+\theta_1k)\right]\cr \ &\eqf{(3)}\ \pdiff{}{x} \pdiff{f}{y}(x_0+\theta_2 h,y_0+\theta_1k)\cr F(h,k)\ &\eqf{(4)}\ \frac{1}{k} \left[\pdiff{f}{x}(x_0+\theta_3h,y_0+k) -\pdiff{f}{x}(x_0+\theta_3h,y_0)\right]\cr \ &\eqf{(5)}\ \pdiff{}{y} \pdiff{f}{x}(x_0+\theta_3 h,y_0+\theta_4k)\cr \end{align*}
for some numbers \(0 \lt \theta_1,\theta_2,\theta_3,\theta_4 \lt 1\text{.}\) All of the numbers \(\theta_1,\theta_2,\theta_3,\theta_4\) depend on \(x_0,y_0,h,k\text{.}\) Hence
\begin{equation*} \pdiff{}{x} \pdiff{f}{y}(x_0+\theta_2 h,y_0+\theta_1k) =\pdiff{}{y} \pdiff{f}{x}(x_0+\theta_3 h,y_0+\theta_4k) \end{equation*}
for all \(h\) and \(k\text{.}\) Taking the limit \((h,k)\rightarrow(0,0)\) and using the assumed continuity of both partial derivatives at \((x_0,y_0)\) gives
\begin{equation*} \lim_{(h,k)\rightarrow (0,0)} F(h,k) =\pdiff{}{x} \pdiff{f}{y}(x_0,y_0) =\pdiff{}{y} \pdiff{f}{x}(x_0,y_0) \end{equation*}
as desired. To complete the proof we just have to justify the details (1), (2), (3), (4) and (5).

Subsubsection 2.3.1.2 The Details

  1. By definition,
    \begin{align*} &\pdiff{}{y} \pdiff{f}{x}(x_0,y_0) =\lim_{k\rightarrow 0}\frac{1}{k} \left[\pdiff{f}{x}(x_0,y_0+k) -\pdiff{f}{x}(x_0,y_0)\right]\cr &=\lim_{k\rightarrow 0}\frac{1}{k} \left[\lim_{h\rightarrow 0}\frac{f(x_0+h,y_0+k)-f(x_0,y_0+k)}{h} -\lim_{h\rightarrow 0}\frac{f(x_0+h,y_0)-f(x_0,y_0)}{h}\right]\cr &=\lim_{k\rightarrow 0}\lim_{h\rightarrow 0} \frac{f(x_0+h,y_0+k)-f(x_0,y_0+k)-f(x_0+h,y_0)+f(x_0,y_0)}{hk}\cr &= \lim_{k\rightarrow 0}\lim_{h\rightarrow 0}F(h,k) \end{align*}
    Similarly,
    \begin{align*} &\pdiff{}{x} \pdiff{f}{y}(x_0,y_0) =\lim_{h\rightarrow 0}\frac{1}{h} \left[\pdiff{f}{y}(x_0+h,y_0) -\pdiff{f}{y}(x_0,y_0)\right]\cr &=\lim_{h\rightarrow 0}\frac{1}{h} \left[\lim_{k\rightarrow 0}\frac{f(x_0+h,y_0+k)-f(x_0+h,y_0)}{k} -\lim_{k\rightarrow 0}\frac{f(x_0,y_0+k)-f(x_0,y_0)}{k}\right]\cr &=\lim_{h\rightarrow 0}\lim_{k\rightarrow 0} \frac{f(x_0+h,y_0+k)-f(x_0+h,y_0)-f(x_0,y_0+k)+f(x_0,y_0)}{hk}\cr &= \lim_{h\rightarrow 0}\lim_{k\rightarrow 0}F(h,k) \end{align*}
  2. The mean value theorem (Theorem 2.13.4 in the CLP-1 text) says that, for any differentiable function \(\varphi(x)\text{,}\)
    • the slope of the line joining the points \(\big(x_0,\varphi(x_0)\big)\) and \(\big(x_0+k,\varphi(x_0+k)\big)\) on the graph of \(\varphi\)
    is the same as
    • the slope of the tangent to the graph at some point between \(x_0\) and \(x_0+k\text{.}\)
    That is, there is some \(0 \lt \theta_1 \lt 1\) such that
    \begin{equation*} \frac{\varphi(x_0+k)-\varphi(x_0)}{k}=\frac{d\varphi}{dx}(x_0+\theta_1 k) \end{equation*}
    Applying this with \(x\) replaced by \(y\) and \(\varphi\) replaced by \(G(y)=f(x_0+h,y)-f(x_0,y)\) gives
    \begin{align*} \frac{G(y_0+k)-G(y_0)}{k} &=\diff{G}{y}(y_0+\theta_1 k) \qquad\text{for some } 0 \lt \theta_1 \lt 1\\ &=\pdiff{f}{y}(x_0+h,y_0+\theta_1k) -\pdiff{f}{y}(x_0,y_0+\theta_1k) \end{align*}
    Hence, for some \(0 \lt \theta_1 \lt 1\text{,}\)
    \begin{align*} F(h,k)\ &=\ \frac{1}{h} \left[\frac{G(y_0+k)-G(y_0)}{k}\right]\\ &=\frac{1}{h} \left[\pdiff{f}{y}(x_0+h,y_0+\theta_1k) -\pdiff{f}{y}(x_0,y_0+\theta_1k)\right] \end{align*}
  3. Define \(H(x)=\pdiff{f}{y}(x,y_0+\theta_1k)\text{.}\) By the mean value theorem,
    \begin{align*} F(h,k)\ &=\ \frac{1}{h}\left[H(x_0+h)-H(x_0)\right]\\ &=\ \diff{H}{x}(x_0+\theta_2 h) \qquad\text{for some } 0 \lt \theta_2 \lt 1\\ &=\pdiff{}{x} \pdiff{f}{y}(x_0+\theta_2 h,y_0+\theta_1k) \end{align*}
  4. Define \(A(x)=f(x,y_0+k)-f(x,y_0)\text{.}\) By the mean value theorem,
    \begin{align*} F(h,k)\ &=\ \frac{1}{k} \left[\frac{A(x_0+h)-A(x_0)}{h}\right]\\ &=\ \frac{1}{k}\diff{A}{x}(x_0+\theta_3 h) \qquad\text{for some } 0 \lt \theta_3 \lt 1\\ &=\frac{1}{k} \left[\pdiff{f}{x}(x_0+\theta_3h,y_0+k) -\pdiff{f}{x}(x_0+\theta_3h,y_0)\right] \end{align*}
  5. Define \(B(y)=\pdiff{f}{x}(x_0+\theta_3h,y)\text{.}\) By the mean value theorem
    \begin{align*} F(h,k)\ &=\ \frac{1}{k}\left[B(y_0+k)-B(y_0)\right]\\ &=\ \diff{B}{y}(y_0+\theta_4 k) \qquad\text{for some } 0 \lt \theta_4 \lt 1\\ &=\pdiff{}{y} \pdiff{f}{x}(x_0+\theta_3 h,y_0+\theta_4k) \end{align*}
This completes the proof of Theorem 2.3.4.

Subsection 2.3.2 Optional — An Example of \(\frac{\partial^2\ f}{\partial x\partial y}(x_0,y_0) \ne\frac{\partial^2\ f}{\partial y\partial x}(x_0,y_0)\)

In Theorem 2.3.4, we showed that \(\frac{\partial^2 f }{\partial x\partial y}(x_0,y_0) =\frac{\partial^2 f }{\partial y\partial x}(x_0,y_0)\) if the partial derivatives \(\frac{\partial^2 f }{\partial x\partial y}\) and \(\frac{\partial^2 f }{\partial y\partial x}\) exist and are continuous at \((x_0,y_0)\text{.}\) Here is an example which shows that if the partial derivatives \(\frac{\partial^2 f }{\partial x\partial y}\) and \(\frac{\partial^2 f }{\partial y\partial x}\) are not continuous at \((x_0,y_0)\text{,}\) then it is possible that \(\frac{\partial^2 f }{\partial x\partial y}(x_0,y_0) \ne\frac{\partial^2 f }{\partial y\partial x}(x_0,y_0)\text{.}\)
Define
\begin{equation*} f(x,y)=\begin{cases} xy\frac{x^2-y^2}{x^2+y^2} & \text{if } (x,y)\ne (0,0) \\ 0 & \text{if } (x,y)=(0,0) \end{cases} \end{equation*}
This function is continuous everywhere. Note that \(f(x,0)=0\) for all \(x\) and \(f(0,y)=0\) for all \(y\text{.}\) We now compute the first order partial derivatives. For \((x,y)\ne (0,0)\text{,}\)
\begin{align*} \pdiff{f}{x}(x,y) &= {\color{blue}{y\frac{x^2-y^2}{x^2+y^2}}} +xy\frac{2x}{x^2+y^2} - xy\frac{2x(x^2-y^2)}{{(x^2+y^2)}^2}\\ &\ = {\color{blue}{y\frac{x^2-y^2}{x^2+y^2}}} + xy\frac{4xy^2}{{(x^2+y^2)}^2}\\ \pdiff{f}{y}(x,y) &= {\color{blue}{x\frac{x^2-y^2}{x^2+y^2}}} -xy\frac{2y}{x^2+y^2} - xy\frac{2y(x^2-y^2)}{{(x^2+y^2)}^2}\\ &\ = {\color{blue}{x\frac{x^2-y^2}{x^2+y^2}}} - xy\frac{4yx^2}{{(x^2+y^2)}^2} \end{align*}
For \((x,y)= (0,0)\text{,}\)
\begin{alignat*}{2} \pdiff{f}{x}(0,0) &= \left[\diff{}{x}f(x,0)\right]_{x=0} &= \left[\diff{}{x} 0\right]_{x=0} &=0\\ \pdiff{f}{y}(0,0) &= \left[\diff{}{y}f(0,y)\right]_{y=0} &= \left[\diff{}{y} 0\right]_{y=0} &=0 \end{alignat*}
By way of summary, the two first order partial derivatives are
\begin{align*} f_x(x,y)&=\begin{cases} y\frac{x^2-y^2}{x^2+y^2} + \frac{4x^2y^3}{{(x^2+y^2)}^2} & \text{if } (x,y)\ne (0,0)\\ 0 & \text{if } (x,y)=(0,0) \end{cases}\\ f_y(x,y)&=\begin{cases} x\frac{x^2-y^2}{x^2+y^2} - \frac{4x^3y^2}{{(x^2+y^2)}^2} & \text{if } (x,y)\ne (0,0)\\ 0 & \text{if } (x,y)=(0,0) \end{cases} \end{align*}
Both \(\pdiff{f}{x}(x,y)\) and \(\pdiff{f}{y}(x,y)\) are continuous. Finally, we compute
\begin{align*} \frac{\partial^2\ f}{\partial x\partial y}(0,0) &=\left[\diff{}{x} f_y(x,0)\right]_{x=0} =\lim_{h\rightarrow 0}\frac{1}{h}\left[f_y(h,0)-f_y(0,0)\right]\\ &=\lim_{h\rightarrow 0}\frac{1}{h}\left[h\frac{h^2-0^2}{h^2+0^2}-0\right] =1\\ \frac{\partial^2\ f}{\partial y\partial x}(0,0) &=\left[\diff{}{y} f_x(0,y)\right]_{y=0} =\lim_{k\rightarrow 0}\frac{1}{k}\left[f_x(0,k)-f_x(0,0)\right]\\ &=\lim_{k\rightarrow 0}\frac{1}{k}\left[k\frac{0^2-k^2}{0^2+k^2}-0\right] =-1 \end{align*}

Exercises 2.3.3 Exercises

Exercise Group.

Exercises — Stage 1
1.
Let all of the third order partial derivatives of the function \(f(x,y,z)\) exist and be continuous. Show that
\begin{align*} f_{xyz}(x,y,z) &=f_{xzy}(x,y,z) =f_{yxz}(x,y,z) =f_{yzx}(x,y,z)\\ &=f_{zxy}(x,y,z) =f_{zyx}(x,y,z) \end{align*}
2.
Find, if possible, a function \(f(x,y)\) for which \(f_x(x,y)=e^y\) and \(f_y(x,y)=e^x\text{.}\)

Exercise Group.

Exercises — Stage 2
3.
Find the specified partial derivatives.
  1. \(f(x,y) = x^2y^3\text{;}\) \(f_{xx}(x,y)\text{,}\) \(f_{xyy}(x,y)\text{,}\) \(f_{yxy}(x,y)\)
  2. \(f(x,y) = e^{xy^2}\text{;}\) \(f_{xx}(x,y)\text{,}\) \(f_{xy}(x,y)\text{,}\) \(f_{xxy}(x,y)\text{,}\) \(f_{xyy}(x,y)\)
  3. \(\displaystyle f(u,v,w) = \frac{1}{u+2v+3w}\ \text{,}\) \(\displaystyle \frac{\partial^3 f}{\partial u\partial v\partial w}(u,v,w)\ \text{,}\) \(\displaystyle \frac{\partial^3 f}{\partial u\partial v\partial w}(3,2,1)\)
4.
Find all second partial derivatives of \(f(x,y)=\sqrt{x^2+5y^2}\text{.}\)
5.
Find the specified partial derivatives.
  1. \(f(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)\text{;}\) \(f_{xyz}(x,y,z)\)
  2. \(f(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big) +\arctan\big(e^{\sqrt{xz}}\big) +\arctan\big(e^{\sqrt{yz}}\big)\text{;}\) \(f_{xyz}(x,y,z)\)
  3. \(f(x,y,z) = \arctan\big(e^{\sqrt{xyz}}\big)\text{;}\) \(f_{xx}(1,0,0)\)
6. (✳).
Let \(f(r,\theta)=r^m\cos m\theta\) be a function of \(r\) and \(\theta\text{,}\) where \(m\) is a positive integer.
  1. Find the second order partial derivatives \(f_{rr}\text{,}\) \(f_{r\theta}\text{,}\) \(f_{\theta\theta}\) and evaluate their respective values at \((r,\theta)=(1,0)\text{.}\)
  2. Determine the value of the real number \(\la\) so that \(f(r,\theta)\) satisfies the differential equation
    \begin{equation*} f_{rr}+\frac{\la}{r}f_r+\frac{1}{r^2}f_{\theta\theta}=0 \end{equation*}

Exercise Group.

Exercises — Stage 3
7.
Let \(\al \gt 0\) be a constant. Show that \(\displaystyle u(x,y,z,t) =\frac{1}{t^{3/2}} e^{-(x^2+y^2+z^2)/(4\al t)}\) satisfies the heat equation
\begin{equation*} u_t = \al\big(u_{xx} + u_{yy} + u_{zz} \big) \end{equation*}
for all \(t \gt 0\)
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