# 4.10 More on partial derivatives  (Page 2/3)

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Suppose $S$ is a subset of ${R}^{2},$ and that $f$ is a continuous real-valued 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 ${C}^{\infty }\left(S\right).$

1. Suppose $f$ is a real-valued function of two real variables and that it is differentiable, as a function of two real variables, at the point $\left(a,b\right).$ Show that the numbers ${L}_{1}$ and ${L}_{2}$ in the definition are exactly the partial derivatives of $f$ at $\left(a,b\right).$ That is,
${L}_{1}=\frac{tialf}{tialx}\left(a,b\right)=\underset{h\to 0}{lim}\frac{f\left(a+h,b\right)-f\left(a,b\right)}{h}$
and
${L}_{2}=\frac{tialf}{tialy}\left(a,b\right)=\underset{h\to 0}{lim}\frac{f\left(a,b+h\right)-f\left(a,b\right)}{h}.$
2. Define $f$ on ${R}^{2}$ as follows: $f\left(0,0\right)=0,$ and if $\left(x,y\right)\ne \left(0,0\right),$ then $f\left(x,y\right)=xy/\left({x}^{2}+{y}^{2}\right).$ Show that both partial derivatives of $f$ at $\left(0,0\right)$ exist and are 0. Show also that $f$ is not , as a function of two real variables, differentiable at $\left(0,0\right).$ HINT: Let $h$ and $k$ run through the numbers $1/n.$
3. What do parts (a) and (b) tell about the relationship between a function of two real variables being differentiable at a point $\left(a,b\right)$ and its having both partial derivatives exist at $\left(a,b\right)?$
4. Suppose $f=u+iv$ is a complex-valued function of a complex variable, and assume that $f$ is differentiable, as a function of a complex variable, at a point $c=a+bi\equiv \left(a,b\right).$ Prove that the real and imaginary parts $u$ and $v$ of $f$ are differentiable, as functions of two real variables. Relate the five quantities
$\frac{tialu}{tialx}\left(a,b\right),\frac{tialu}{tialy}\left(a,b\right),\frac{tialv}{tialx}\left(a,b\right),\frac{tialv}{tialy}\left(a,b\right),\phantom{\rule{4.pt}{0ex}}\text{and}\phantom{\rule{4.pt}{0ex}}{f}^{\text{'}}\left(c\right).$

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 $\left(a,b\right)$ of the interior of $S,$ and assume in addition that one of these second order partials exists at every point in a disk ${B}_{r}\left(a,b\right)$ around $\left(a,b\right)$ and that it is continuous at the point $\left(a,b\right).$ Then ${f}_{xy}\left(a,b\right)={f}_{yx}\left(a,b\right).$

Suppose that it is ${f}_{yx}$ that is continuous at $\left(a,b\right).$ Let $ϵ>0$ be given, and let ${\delta }_{1}>0$ be such that if $|\left(c,d\right)-\left(a,b\right)|<{\delta }_{1}$ then $|{f}_{yx}\left(c,d\right)-{f}_{yx}\left(a,b\right)|<ϵ.$ Next, choose a ${\delta }_{2}$ such that if $0<|k|<{\delta }_{2},$ then

$|{f}_{xy}\left(a,b\right)-\frac{{f}_{x}\left(a,b+k\right)-{f}_{x}\left(a,b\right)}{k}|<ϵ,$

and fix such a $k.$ We may also assume that $|k|<{\delta }_{1}/2.$ Finally, choose a ${\delta }_{3}>0$ such that if $0<|h|<{\delta }_{3},$ then

$|{f}_{x}\left(a,b+k\right)-\frac{f\left(a+h,b+k\right)-f\left(a,b+k\right)}{h}|<|k|ϵ,$

and

$|{f}_{x}\left(a,b\right)-\frac{f\left(a+h,b\right)-f\left(a,b\right)}{h}|<|k|ϵ,$

and fix such an $h.$ Again, we may also assume that $|h|<{\delta }_{1}/2.$

In the following calculation we will use the Mean Value Theorem twice.

$\begin{array}{ccc}\hfill 0& \le & |{f}_{xy}\left(a,b\right)-{f}_{yx}\left(a,b\right)|\hfill \\ & \le & |{f}_{xy}\left(a,b\right)-\frac{{f}_{x}\left(a,b+k\right)-{f}_{x}\left(a,b\right)}{k}|\hfill \\ & & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}+|\frac{{f}_{x}\left(a,b+k\right)-{f}_{x}\left(a,b\right)}{k}-{f}_{yx}\left(a,b\right)|\hfill \\ & \le & ϵ+|\frac{{f}_{x}\left(a,b+k\right)-\frac{f\left(a+h,b+k\right)-f\left(a,b+k\right)}{h}}{k}|\hfill \\ & & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}+|\frac{\frac{f\left(a+h,b\right)-f\left(a,b\right)}{h}-{f}_{x}\left(a,b\right)}{k}|\hfill \\ & & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}+|\frac{f\left(a+h,b+k\right)-f\left(a,b+k\right)+\left(f\left(a+h,b\right)-f\left(a,b\right)\right)}{hk}-{f}_{yx}\left(a,b\right)|\hfill \\ & <& 3ϵ+|\frac{f\left(a+h,b+k\right)-f\left(a,b+k\right)+\left(f\left(a+h,b\right)-f\left(a,b\right)\right)}{hk}-{f}_{yx}\left(a,b\right)|\hfill \\ & =& 3ϵ+|\frac{{f}_{y}\left(a+h,{b}^{\text{'}}\right)-{f}_{y}\left(a,{b}^{\text{'}}\right)}{h}-{f}_{yx}\left(a,b\right)|\hfill \\ & =& 3ϵ+|{f}_{yx}\left({a}^{\text{'}},{b}^{\text{'}}\right)-{f}_{yx}\left(a,b\right)|\hfill \\ & <& 4ϵ,\hfill \end{array}$

because ${b}^{\text{'}}$ is between $b$ and $b+k,$ and ${a}^{\text{'}}$ is between $a$ and $a+h,$ so that $|\left({a}^{\text{'}},{b}^{\text{'}}\right)-\left(a,b\right)|<{\delta }_{1}/\sqrt{2}<{\delta }_{1}.$ Hence, $|{f}_{xy}\left(a,b\right)-{f}_{yx}\left(a,b\right)<4ϵ,$ for an arbitrary $ϵ,$ and so the theorem is proved.

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