
Suppose
$S$ is a subset of
${R}^{2},$ and that
$f$ is a continuous realvalued function on
$S.$ If both partial derivatives of
$f$ exist at each point of the interior
${S}^{0}$ of
$S,$ and both
are continuous on
${S}^{0},$ then
$f$ is said to belong to
${C}^{1}\left(S\right).$ If all
$k$ th order mixed partial derivatives exist at each point of
${S}^{0},$ and all of them are continuous on
${S}^{0},$ then
$f$ is said to belong to
${C}^{k}\left(S\right).$ Finally, if all mixed partial derivatives, of arbitrary orders, exist and are continuous on
${S}^{0},$ then
$f$ is said to belong to
${C}^{\infty}\left(S\right).$
 Suppose
$f$ is a realvalued function of
two real variables and that it is differentiable,
as a function of two real variables, at the point
$(a,b).$ Show that the numbers
${L}_{1}$ and
${L}_{2}$ in the definition are exactly the
partial derivatives of
$f$ at
$(a,b).$ That is,
$${L}_{1}=\frac{tialf}{tialx}(a,b)=\underset{h\to 0}{lim}\frac{f(a+h,b)f(a,b)}{h}$$
and
$${L}_{2}=\frac{tialf}{tialy}(a,b)=\underset{h\to 0}{lim}\frac{f(a,b+h)f(a,b)}{h}.$$
 Define
$f$ on
${R}^{2}$ as follows:
$f(0,0)=0,$ and if
$(x,y)\ne (0,0),$ then
$f(x,y)=xy/({x}^{2}+{y}^{2}).$ Show that both partial derivatives of
$f$ at
$(0,0)$ exist and are 0.
Show also that
$f$ is
not , as a function of two real variables, differentiable at
$(0,0).$ HINT: Let
$h$ and
$k$ run through the numbers
$1/n.$
 What do parts (a) and (b) tell about the relationship between a function of two real variables
being differentiable at a point
$(a,b)$ and its having
both partial derivatives exist at
$(a,b)?$
 Suppose
$f=u+iv$ is a complexvalued function of a complex variable,
and assume that
$f$ is differentiable, as a function of a complex variable, at a point
$c=a+bi\equiv (a,b).$ Prove that the real and imaginary parts
$u$ and
$v$ of
$f$ are differentiable, as functions of two real variables.
Relate the five quantities
$$\frac{tialu}{tialx}(a,b),\frac{tialu}{tialy}(a,b),\frac{tialv}{tialx}(a,b),\frac{tialv}{tialy}(a,b),\phantom{\rule{4.pt}{0ex}}\text{and}\phantom{\rule{4.pt}{0ex}}{f}^{\text{'}}\left(c\right).$$
Perhaps the most interesting theorem about partial derivatives is the
“mixed partials are equal” theorem. That is,
${f}_{xy}={f}_{yx}.$ The point is that this is
not always the case. An extra
hypothesis is necessary.
Theorem on mixed partials
Let
$f:S\to R$ be such that both second order partials derivatives
${f}_{xy}$ and
${f}_{yx}$ exist at a point
$(a,b)$ of the interior of
$S,$ and assume in
addition that one of these second order partials exists at every point in a disk
${B}_{r}(a,b)$ around
$(a,b)$ and that it is continuous at the point
$(a,b).$ Then
${f}_{xy}(a,b)={f}_{yx}(a,b).$
Suppose that it is
${f}_{yx}$ that is continuous at
$(a,b).$ Let
$\u03f5>0$ be given, and let
${\delta}_{1}>0$ be such that
if
$(c,d)(a,b)<{\delta}_{1}$ then
${f}_{yx}(c,d){f}_{yx}(a,b)<\u03f5.$ Next, choose a
${\delta}_{2}$ such that if
$0<\leftk\right<{\delta}_{2},$ then
$${f}_{xy}(a,b)\frac{{f}_{x}(a,b+k){f}_{x}(a,b)}{k}<\u03f5,$$
and fix such a
$k.$ We may also assume that
$\leftk\right<{\delta}_{1}/2.$ Finally, choose a
${\delta}_{3}>0$ such that if
$0<\lefth\right<{\delta}_{3},$ then
$${f}_{x}(a,b+k)\frac{f(a+h,b+k)f(a,b+k)}{h}<k\u03f5,$$
and
$${f}_{x}(a,b)\frac{f(a+h,b)f(a,b)}{h}<k\u03f5,$$
and fix such an
$h.$ Again, we may also assume that
$\lefth\right<{\delta}_{1}/2.$
In the following calculation we will use the Mean Value Theorem twice.
$$\begin{array}{ccc}\hfill 0& \le & {f}_{xy}(a,b){f}_{yx}(a,b)\hfill \\ & \le & {f}_{xy}(a,b)\frac{{f}_{x}(a,b+k){f}_{x}(a,b)}{k}\hfill \\ & & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}+\frac{{f}_{x}(a,b+k){f}_{x}(a,b)}{k}{f}_{yx}(a,b)\hfill \\ & \le & \u03f5+\frac{{f}_{x}(a,b+k)\frac{f(a+h,b+k)f(a,b+k)}{h}}{k}\hfill \\ & & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}+\frac{\frac{f(a+h,b)f(a,b)}{h}{f}_{x}(a,b)}{k}\hfill \\ & & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}+\frac{f(a+h,b+k)f(a,b+k)+\left(f\right(a+h,b)f(a,b\left)\right)}{hk}{f}_{yx}(a,b)\hfill \\ & <& 3\u03f5+\frac{f(a+h,b+k)f(a,b+k)+\left(f\right(a+h,b)f(a,b\left)\right)}{hk}{f}_{yx}(a,b)\hfill \\ & =& 3\u03f5+\frac{{f}_{y}(a+h,{b}^{\text{'}}){f}_{y}(a,{b}^{\text{'}})}{h}{f}_{yx}(a,b)\hfill \\ & =& 3\u03f5+{f}_{yx}({a}^{\text{'}},{b}^{\text{'}}){f}_{yx}(a,b)\hfill \\ & <& 4\u03f5,\hfill \end{array}$$
because
${b}^{\text{'}}$ is between
$b$ and
$b+k,$ and
${a}^{\text{'}}$ is between
$a$ and
$a+h,$ so that
$({a}^{\text{'}},{b}^{\text{'}})(a,b)<{\delta}_{1}/\sqrt{2}<{\delta}_{1}.$ Hence,
${f}_{xy}(a,b){f}_{yx}(a,b)<4\u03f5,$ for an arbitrary
$\u03f5,$ and so the theorem is proved.