Let
$f$ be defined on
${R}^{2}$ by
$f(0,0)=0$ and, for
$(x,y)\ne (0,0),$
$f(x,y)={x}^{3}y/({x}^{2}+{y}^{2}).$
- Prove that both partial derivatives
${f}_{x}$ and
${f}_{y}$ exist
at each point in the plane.
- Show that
${f}_{yx}(0,0)=1$ and
${f}_{xy}(0,0)=0.$
- Show that
${f}_{xy}$ exists at each point in the plane, but that it is not continuous at
$(0,0).$
The following exercise is an obvious generalization of the First Derivative Test for Extreme Values,
[link] , to real-valued functions of two real variables.
Let
$f:S\to R$ be a real-valued function of two real variables,
and let
$c=(a,b)\in {S}^{0}$ be a point at which
$f$ attains a local maximum or a local minimum.
Show that ifeither of the partial derivatives
$tialf/tialx$ or
$tialf/tialy$ exists at
$c,$ then it must be equal to 0.
HINT: Just consider real-valued functions of a real variable like
$x\to f(x,b)$ or
$y\to f(a,y),$ and use
[link] .
Whenever we make a new definition about functions, the question arisesof how the definition fits with algebraic combinations of
functions and how it fits with the operation of composition.In that light, the next theorem is an expected one.
(Chain Rule again)
Suppose
$S$ is a subset of
${R}^{2},$ that
$(a,b)$ is a point in the interior of
$S,$ and that
$f:S\to R$ is a real-valued function that is differentiable, as a function of two real variables, at the point
$(a,b).$ Suppose that
$T$ is a subset of
$R,$ that
$c$ belongs to the interior of
$T,$ and that
$\phi :T\to {R}^{2}$ is differentiable at the point
$c$ and
$\phi \left(c\right)=(a,b).$ Write
$\phi \left(t\right)=\left(x\right(t),y(t\left)\right).$ Then the composition
$f\circ \phi $ is differentiable at
$c$ and
$$f\circ {\phi}^{\text{'}}\left(c\right)=\frac{tialf}{tialx}(a,b){x}^{\text{'}}\left(c\right)+\frac{tialf}{tialy}(a,b){y}^{\text{'}}\left(c\right)=\frac{tialf}{tialx}\left(\phi \left(c\right)\right){x}^{\text{'}}\left(c\right)+\frac{tialf}{tialy}\left(\phi \left(c\right)\right){y}^{\text{'}}\left(c\right).$$
From the definition of differentiability of a real-valued function of two real variables, write
$$f(a+{h}_{1},b+{h}_{2})-f(a,b)={L}_{1}{h}_{1}+{L}_{2}{h}_{2}+{\theta}_{f}({H}_{1},{h}_{2}).$$
and from part (3) of
[link] , write
$$\phi (c+h)-\phi \left(c\right)={\phi}^{\text{'}}\left(c\right)h+{\theta}_{\phi}\left(h\right),$$
or, in component form,
$$x(c+h)-x\left(c\right)=x(c+h)-a={x}^{\text{'}}\left(c\right)h+{\theta}_{x}\left(h\right)$$
and
$$y(c+h)-y\left(c\right)=y(c+h)-b={y}^{\text{'}}\left(c\right)h+{\theta}_{y}\left(h\right).$$
We also have that
$$\underset{\left|\right({h}_{1},{h}_{2}\left)\right|\to 0}{lim}\frac{{\theta}_{f}\left(({h}_{1},{h}_{2})\right)}{\left|\right({h}_{1},{h}_{2}\left)\right|}=0,$$
$$\underset{h\to 0}{lim}\frac{{\theta}_{x}\left(h\right)}{h}=0,$$
and
$$\underset{h\to 0}{lim}\frac{{\theta}_{y}\left(h\right)}{h}=0.$$
We will show that
$f\circ \phi $ is differentiable at
$c$ by showing
that there exists a number
$L$ and a function
$\theta $ satisfying
the two conditions of part (3) of
[link] .
Define
$${k}_{1}\left(h\right),{k}_{2}\left(h\right))=\phi (c+h)-\phi \left(c\right)=(x(c+h)-x\left(c\right),y(c+h)-y\left(c\right)).$$
Thus, we have that
$$\begin{array}{ccc}\hfill f\circ \phi (c+h)-f\circ \phi \left(c\right)& =& f\left(\phi \right(c+h\left)\right)-f\left(\phi \right(c\left)\right)\hfill \\ & =& f\left(x\right(c+h),y(c+h\left)\right)-f\left(x\right(c),y(c\left)\right)\hfill \\ & =& f(a+{k}_{1}\left(h\right),b+{k}_{2}\left(h\right))-f(a,b)\hfill \\ & =& {L}_{1}{k}_{1}\left(h\right)+{L}_{2}{k}_{2}\left(h\right)+{\theta}_{f}({k}_{1}\left(h\right),{k}_{2}\left(h\right))\hfill \\ & =& {l}_{1}(x(c+h)-x\left(c\right))+{L}_{2}(y(c+h)-y\left(c\right))\hfill \\ & & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}+{\theta}_{f}({k}_{1}\left(h\right),{k}_{2}\left(h\right))\hfill \\ & =& {L}_{1}({x}^{\text{'}}\left(c\right)h+{\theta}_{x}\left(h\right))+{L}_{2}({y}^{\text{'}}\left(c\right)h+{\theta}_{y}\left(h\right))\hfill \\ & & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}+{\theta}_{f}({k}_{1}\left(h\right),{k}_{2}\left(h\right))\hfill \\ & =& ({L}_{1}{x}^{\text{'}}\left(c\right)+{L}_{2}{y}^{\text{'}}\left(c\right))h\hfill \\ & & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}+{L}_{1}{\theta}_{x}\left(h\right)+{L}_{2}{\theta}_{y}\left(h\right)+{\theta}_{f}({k}_{1}\left(h\right),{k}_{2}\left(h\right)).\hfill \end{array}$$
We define
$L=({L}_{1}{x}^{\text{'}}\left(c\right)+{L}_{2}{y}^{\text{'}}\left(c\right))$ and
$\theta \left(h\right)={l}_{1}{\theta}_{x}\left(h\right)+{L}_{2}{\theta}_{y}\left(h\right)+{\theta}_{f}({k}_{1}\left(h\right),{k}_{2}\left(h\right)).$ By these definitions and the calculation above we have Equation (4.1)
$$f\circ \phi (c+h)-f\circ \phi \left(c\right)=Lh+\theta \left(h\right),$$
so that it only remains to verify
[link] for the function
$\theta .$ We have seen above that the first two parts of
$\theta $ satisfy the desired limit condition,
so that it is just the third part of
$\theta $ that requires some proof.
The required argument is analogous to the last part of the proof of the Chain Rule (
[link] ),
and we leave it as an exercise.
- Finish the proof to the preceding theorem by showing that
$$\underset{h\to 0}{lim}\frac{{\theta}_{f}({k}_{1}\left(h\right),{k}_{2}\left(h\right))}{h}=0.$$
HINT: Review the corresponding part of the proof to
[link] .
- Suppose
$f:S\to R$ is as in the preceding theorem and that
$\phi $ is a
real-valued function of a real variable.Suppose
$f$ is differentiable, as a function of two real variables,
at the point
$(a,b),$ and that
$\phi $ is differentiable
at the point
$c=f(a,b).$ Let
$g=\phi \circ f.$ Find a formula for the partial derivatives of the
real-valued function
$g$ of two real variables.
- (A generalized Mean Value Theorem)
Suppose
$u$ is a real-valued function of two real variables,
both of whose partial derivatives exist at each point in a disk
${B}_{r}(a,b).$ Show that, for any two points
$(x,y)$ and
$({x}^{\text{'}},{y}^{\text{'}})$ in
${B}_{r}(a,b),$ there exists a point
$(\widehat{x},\widehat{y})$ on the line segment joining
$(x,y)$ to
$({x}^{\text{'}},{y}^{\text{'}})$ such that
$$u(x,y)-u({x}^{\text{'}},{y}^{\text{'}})=\frac{tialu}{tialx}(\widehat{x},\widehat{y})(x-{x}^{\text{'}})+\frac{tialu}{tialy}(\widehat{x},\widehat{y})(y-{y}^{\text{'}}).$$
HINT: Let
$\phi :[0,1]\to {R}^{2}$ be defined by
$\phi \left(t\right)=(1-t)({x}^{\text{'}},{y}^{\text{'}})+t(x,y).$ Now use the preceding theorem.
- Verify that the assignment
$f\to tialf/tialx$ is linear; i.e., that
$$\frac{tial(f+g)}{tialx}=\frac{tialf}{tialx}+\frac{tialg}{tialx}.$$
Check that the same is true for partial derivatives with respect to
$y.$