Home Page Dirk Bollaerts & Céleste Paalvast

1. Examples of exercises

Two completely solved exercises

Exercise 1

Discuss the

continuity,
partial derivatives,
directional derivatives,
differentiability,
continuity of the partial derivatives

of the following function in

(0,0)

$f (x, y) = {\begin{matrix} \frac{3 x^{2} y - y^{3}}{x^{2} + y^{2}} & if (x, y) \neq (0,0), \\ 0 & if (x, y) = (0,0) . \end{matrix}$

Solution

1. Continuity of the function

We investigate continuity with an $ϵ$ - $δ$ approach.

When applying our function definition, we have then the following statements. Try to find a $δ$ such that if $| (x, y) - (0,0) | = \sqrt{x^{2} + y^{2}} < δ$ , we have that

$| \frac{3 x^{2} y - y^{3}}{x^{2} + y^{2}} - 0 | < ϵ .$

$\begin{aligned} | \frac{3 x^{2} y - y^{3}}{x^{2} + y^{2}} | & \leq \frac{3 x^{2} | y | + | y |^{3}}{x^{2} + y^{2}} \\ \leq \frac{3 {\sqrt{x^{2} + y^{2}}}^{2} \sqrt{x^{2} + y^{2}} + {\sqrt{x^{2} + y^{2}}}^{3}}{{\sqrt{x^{2} + y^{2}}}^{2}} \\ \leq \frac{4 {\sqrt{x^{2} + y^{2}}}^{3}}{{\sqrt{x^{2} + y^{2}}}^{2}} \\ \leq 4 \sqrt{x^{2} + y^{2}} . \end{aligned}$

It is sufficient to take $δ = ϵ ⁄ 4$ . We can find a $δ$ , so we conclude that the function is continuous.

3D View of the surface. — We see here a three dimensional figure of the graph of the function. This looks like a continuous function.

Contourplot of the surface. — We see here a figure of the contour plot of the function. Only level curves of level around $0$ come close to $(0,0)$ .

2. Partial derivatives

Discussion of the partial derivative to x in f(x, y)

We observe then that $f |_{y = 0} (x, y) = {\begin{matrix} f (x, 0) = 0 & i f x \neq 0; \\ 0 & i f x = 0 . \end{matrix}$

So $\begin{aligned} \frac{\partial f}{\partial x} (0,0) & = \lim_{h \to 0} \frac{f (h, 0) - f (0,0)}{h} \\ = \lim_{h \to 0} 0 \\ = 0 . \end{aligned}$ So the partial derivative to $x$ does exist.

Discussion of the partial derivative to y in f(x, y)

We observe that $f |_{x = 0} (x, y) = {\begin{matrix} f (0, y) = - y & if y \neq 0; \\ 0 & if y = 0 . \end{matrix}$

So $\begin{aligned} \frac{\partial f}{\partial y} (0,0) & = \lim_{h \to 0} \frac{f (0, h) - f (0,0)}{h} \\ = \lim_{h \to 0} - 1 \\ = - 1 . \end{aligned}$ So the partial derivative to $y$ does exist.

We conclude that $\frac{\partial f}{\partial x} (0,0) = 0 and \frac{\partial f}{\partial y} (0,0) = - 1 .$

function restricted to the vertical Y-axis. — We see here a figure of the graph of the function restricted to the vertical $Y$ -axis through $(0,0)$ . We have drawn here the function $f (0, h)$ .

3. Directional derivatives

Discussion of the directional derivatives in

(0,0)

Let $(u, v)$ be a normalised direction vector. So it has length $1$ , this means that $\sqrt{u^{2} + v^{2}} = 1$ . Then the directional derivative in $(0,0)$ in the direction $(u, v)$ , notation $D_{(u, v)} (0,0)$ is by definition $D_{(u, v)} (0,0) = \lim_{h \to 0} \frac{f (0 + h u, 0 + h v) - f (0,0)}{h} .$

The classical partial derivatives are of course also directional derivatives in the directions $(1,0)$ and $(0,1)$ .

In the case of our function we have to calculate $\begin{aligned} \lim_{h \to 0} \frac{f (0 + h u, 0 + h v) - f (0,0)}{h} & = \lim_{h \to 0} \frac{3 h^{3} u^{2} v - h^{3} v^{3}}{h (h^{2} u^{2} + h^{2} v^{2})} \\ = - \frac{v (v^{2} - 3 u^{2})}{u^{2} + v^{2}} . \end{aligned}$

$\begin{aligned} D_{(u, v)} (0,0) & = \lim_{h \to 0} \frac{f (0 + h u, 0 + h v) - f (0,0)}{h} \\ = \lim_{h \to 0} - v (v^{2} - 3 u^{2}) \\ = - v (v^{2} - 3 u^{2}) . \end{aligned}$ We have used that $(u, v)$ is a normal vector and so $u^{2} + v^{2} = 1$ .

So the directional derivatives do always exist.

4. Differentiability

Some preliminary visual tests

Before setting out to do an investigation of the differentiability, let us take the liberty to do a few visual checks. Maybe these figures can make us doubtful about the differentiability. Because we do not rely on purely visual proofs, we will then continue our reasoning as if we did not perform these visual checks.

The partial derivatives exist and all other directional derivatives exist. The function is also continuous. So it can be useful to show the function and its candidate tangent plane in $(0,0)$ in order to see what is going on.

3dim view of the function and the tangent plane. — We see here a three dimensional figure of the graph of the function $\frac{\partial f}{\partial x} (0,0) x + \frac{\partial f}{\partial y} (0,0) y$ , which is graphically the candidate tangent plane and the function $f (x, y)$ . We see here that the candidate tangent plane does not fit the function very nicely. It is indeed no tangent plane following our future calculations.

We perform now our second visual check. We can take a look at it in another way. We have calculated all the directional derivatives and we know that if the function is differentiable, then the following vectors $(u, v, D_{(u, v)} (0,0))$ must lie in one plane, which is the tangent plane if the function is differentiable. So let us visually check that these vectors are coplanar.

View of the directional derivatives. — We see here a three dimensional figure of the graph of the function and the position of the vectors $(u, v, D_{(u, v)} (0,0))$ which must sweep out an ellipse in the candidate tangent plane. If they do not, then there is no tangent plane and the function cannot be differentiable. We see the red circle of the unit vectors $(u, v)$ in the $X$ - $Y$ plane. We see also in the blue curve the vectors $(u, v, D_{(u, v)} (0,0))$ . Four of them are indicated by large red points. We see here that the vectors $(u, v, D_{(u, v)} (0,0))$ do not sweep out a nice ellipse lying in the candidate tangent plane! Only four points are in the candidate tangent plane. On this visual basis, we conclude that the function is not differentiable, but we stated that we want to give an alphabetical proof following the definition.

Defining the quotient function

We are going to define the quotient for this function.

We define the quotient function, notation $q (h, k)$ in $(0,0)$ as

$q (h, k) = \frac{f (h, k) - f (0,0) - (\frac{\partial f}{\partial x} (0,0) h + \frac{\partial f}{\partial y} (0,0) k)}{| (h, k) |} .$

Remark that this quotient is the two variable equivalent of the one variable quotient

$q (h) = \frac{f (h) - f (0) - f^{'} (0) h}{h} .$

So please do not confuse this with $\frac{f (h) - f (0)}{h}$ , which is commonly called the differential quotient!

$\lim_{(h, k) \to (0,0)} q (h, k) = 0,$ then the function $f (x, y)$ is by definition differentiable in $(0,0)$ .

So we have to prove that the function

$\begin{matrix} q (h, k) = {\begin{matrix} \frac{4 h^{2} k}{{(h^{2} + k^{2})}^{3 ⁄ 2}} & if (h, k) \neq (0,0); \\ 0 & if (h, k) = (0,0) \end{matrix} \end{matrix}$

is continuous in $(0,0)$ .

Discussion of the continuity of q(h,k) in (0,0)

We restrict the function $q (h, k)$ to the continuous curves with equations $k = λ h$ . We observe then that

$\begin{aligned} q |_{k = λ h} (h, k) \\ = {\begin{matrix} \frac{4 h^{3} λ}{{(h^{2} λ^{2} + h^{2})}^{3 ⁄ 2}} = \frac{4 h^{3} λ}{{(λ^{2} + 1)}^{3 ⁄ 2} | h |^{3}} & if h \neq 0; \\ 0 & if h = 0 . \end{matrix} \end{aligned}$

We see that these restricted functions have no limits. But if $q (h, k)$ is continuous, all these limit values should be $q (0, 0) = 0$ . So this function $q (h, k)$ is not continuous in $(0,0)$ . The function $f (x, y)$ is not differentiable in $(0,0)$ .

Graph of the function q (h, k). — We see here a three dimensional figure of the graph of the function $q (h, k)$ . The vertical line above $(0,0)$ looks suspicious. This does not seem to be a graph of a continuous function.

Function restricted to lines through (0,0). — We have restricted the function $q (h, k)$ here to $k = 1 ⁄ 2 h$ and $k = 3 ⁄ 10 h$ and $k = 9 ⁄ 10 h$ . We see in this figure clearly that the restrictions of the function to these lines are functions that have no limits in $0$ .

contour plot of the function q (h, k). — We see here a figure of the contour plot of the function $q (h, k)$ . Many level curves of very different levels approach $(0,0)$ . This looks discontinuous indeed.

5. Discussion of the continuity of the partial derivatives

It is in this context useless to talk about an alternative proof for differentiability because the function is not differentiable. So this section is irrelevant.

6. Overview and summary

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{3 x^{2} y - y^{3}}{x^{2} + y^{2}} & if (x, y) \neq (0,0), \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

continuity	yes
partial derivatives exist	yes
all directional derivatives exist	yes
differentiable	no
partials are continuous	irrelevant

7. One step further

Possible further investigations

We have used in the calculations for differentiability that we had some magical curves $k = λ h$ which behaved very strangely when mapped by $q (h, k)$ to the $Z$ -direction. We want to see what is going on with these curves. Let us define the 3-dimensional curve in parametric form that projects in the $(h, k)$ -plane to our curve $k = λ h$ : $(x (t), y (t), z (t)) = (t, λ t, f (t, λ t)) = (t, λ t, - \frac{λ (λ^{2} - 3) t}{λ^{2} + 1}) .$

This curve lies completely in the surface defined by the function. It is clear that the tangent vector lies in the tangent plane if the function is differentiable. Now we had a candidate tangent plane, we draw that and see what is going on with the tangent vector $(x^{'} (t), y^{'} (t), z^{'} (t)) = (1, λ, - \frac{λ (λ^{2} - 3)}{λ^{2} + 1}) .$

We also know that the candidate tangent plane is the only possible tangent plane because it takes care of a good fit in the $X$ -direction and the $Y$ -direction, which is the absolute minimum that a tangent plane must do. So if $t = 0$ , then we have the tangent vector $(1, λ, - \frac{λ (λ^{2} - 3)}{λ^{2} + 1}) .$

If $λ = 1$ , this leaves us with the tangent vector $(1,1,1)$ . We will see that this vector does not lie in the tangent plane. So we see once again that the candidate tangent plane is not a real tangent plane. Please consult the figure of this situation.

Tangent vector not lying in the candidate tangent plane. — The tangent vector is not in the only possible candidate tangent plane. Because the curve is in this example a line, we have that this tangent vector is on the line itself. This tangent vector intersects the plane transversally.

7. Alternative proof for the non differentiability

Suppose that we already met in the course the differentiation rule of the composition of two differentiable functions. This is also called the chain rule. Then we have proven the following. If the function is differentiable in $(a, b)$ , then the directional derivative can be calculated as follows.

$D_{(u, v)} f (a, b) = \frac{\partial f}{\partial x} (a, b) u + \frac{\partial f}{\partial y} (a, b) v .$

Important remark. This formula is only valid if the function is differentiable. One of the most common mistakes is that one uses this formula in the case of non differentiability. It seems to be easy to calculate quickly the partial derivatives if they exist and then use this formula.

We have calculated the directional derivatives and we saw that

D_{(u, v)} f (0,0) = - v (v^{2} - 3 u^{2})

and this is certainly not the linear function in $u$ and $v$ which we should have in the case of differentiability.

So we conclude again with this alternative proof that the function is not differentiable.

8. Behaviour of the gradient

We know that if a function has continuous partial derivatives, then the function must be differentiable. It follows that if a function is not differentiable, then the gradient of the function, if it exists, cannot be continuous. The vector field defined by the gradient must behave in the case of non differentiability quite peculiarly. So it can be interesting to take a look at the gradient vector field.

Vector field of the gradient. — We made here the following sketch. We have drawn the graphics of $y = x$ in pink and $y = 0.62 x$ in cyan. We have sketched the gradient vector field $(\frac{\partial f}{\partial x} (x, y), \frac{\partial f}{\partial y} (x, y))$ which is not continuous as can be seen as follows. Observe the gradient vector field on the pink curve, these are the red vectors. Observe the gradient vector field on the cyan curve, these are the green vectors. The purple vector is the gradient vector in $(0, 0)$ . The red vectors converge to a vector with a non zero $x$ -component. This component is equal to $2$ . The green vectors converge to a vector that has a $x$ -component that is two times smaller then the $x$ -component of the vector to which the red vector field converges if $x \to 0$ . This is clearly impossible if the vector field is continuous. Moreover, the $x$ -component of the limit vector should in both cases be zero if the gradient vector field is continuous. We conclude that this gradient vector field is not continuous. The function is differentiable in that case, which it is not. Please note however that a sketch of the gradient vector field is inherently a sketch of discrete data. So the utmost care must be taken in order to make it a little bit trustworthy.

Exercise 2

Discuss the

continuity,
partial derivatives,
directional derivatives,
differentiability,
continuity of the partial derivatives

of the following function in

(0,0)

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x y (x^{2} - y^{2})}{x^{2} + y^{2}} & if (x, y) \neq (0,0), \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Solution

1. Continuity of the function

We investigate continuity with an $ϵ$ - $δ$ approach.

When applying our function definition, we have then the following statements. Try to find a $δ$ such that if $| (x, y) - (0,0) | = \sqrt{x^{2} + y^{2}} < δ$ , we have that

$| \frac{x y (x^{2} - y^{2})}{x^{2} + y^{2}} - 0 | < ϵ .$

We are looking for a function $ℓ (x, y)$ that is larger then or equal to the left hand side of this last inequality. This function $ℓ (x, y)$ has to have the property that it can be made smaller then $ϵ$ by carefully manipulating the value $δ .$ This is sufficient for our continuity proof. $\begin{aligned} | \frac{x y (x^{2} - y^{2})}{x^{2} + y^{2}} | \\ \leq \frac{| x | | y | (| x |^{2} + | y |^{2})}{x^{2} + y^{2}} \\ \leq \frac{\sqrt{x^{2} + y^{2}} \sqrt{x^{2} + y^{2}} ({\sqrt{x^{2} + y^{2}}}^{2} + {\sqrt{x^{2} + y^{2}}}^{2})}{{\sqrt{x^{2} + y^{2}}}^{2}} \\ \leq \frac{2 {\sqrt{x^{2} + y^{2}}}^{4}}{{\sqrt{x^{2} + y^{2}}}^{2}} \\ \leq 2 {\sqrt{x^{2} + y^{2}}}^{2} . \end{aligned}$ It is sufficient to take $δ = (ϵ ⁄ 2)^{1 ⁄ 2}$ . We can find a $δ$ , so we conclude that the function is continuous.

2. Partial derivatives

Discussion of the partial derivative of f(x,y) in x

We observe then that $f |_{y = 0} (x, y) = {\begin{matrix} f (x, 0) = 0 & if x \neq 0; \\ 0 & if x = 0 . \end{matrix}$

So $\begin{aligned} \frac{\partial f}{\partial x} (0,0) & = \lim_{h \to 0} \frac{f (h, 0) - f (0,0)}{h} \\ = \lim_{h \to 0} 0 \\ = 0 . \end{aligned}$ So the partial derivative to $x$ does exist.

Discussion of the partial derivative of f(x,y) in x

We observe that $f |_{x = 0} (x, y) = {\begin{matrix} f (0, y) = 0 & if y \neq 0; \\ 0 & if y = 0 . \end{matrix}$

So $\begin{aligned} \frac{\partial f}{\partial y} (0,0) & = \lim_{h \to 0} \frac{f (0, h) - f (0,0)}{h} \\ = \lim_{h \to 0} 0 \\ = 0 . \end{aligned}$ So the partial derivative to $y$ does exist.

We conclude that $\frac{\partial f}{\partial x} (0,0) = 0 and \frac{\partial f}{\partial y} (0,0) = 0 .$

3. Directional derivatives

Discussion of the directional derivatives in (0,0)

Let $(u, v)$ be a normalised direction vector. So it has length $1$ , this means that
$\sqrt{u^{2} + v^{2}} = 1$ . Then the directional derivative in $(0,0)$ in the direction $(u, v)$ , notation $D_{(u, v)} (0,0)$ is by definition $D_{(u, v)} (0,0) = \lim_{h \to 0} \frac{f (0 + h u, 0 + h v) - f (0,0)}{h} .$

The classical partial derivatives are of course also directional derivatives in the directions $(1,0)$ and $(0,1)$ .

In the case of our function we have to calculate the following limit $\begin{aligned} D_{(u, v)} (0,0) & = \lim_{h \to 0} \frac{f (0 + h u, 0 + h v) - f (0,0)}{h} \\ = \lim_{h \to 0} \frac{h u v (h^{2} u^{2} - h^{2} v^{2})}{h^{2} u^{2} + h^{2} v^{2}} \\ = \lim_{h \to 0} \frac{h u v (u^{2} - v^{2})}{u^{2} + v^{2}} \\ = 0 . \end{aligned}$

So the directional derivatives do always exist.

4. Differentiability

Some preliminary visual tests

The function is continuous and all directional derivatives exist, so there is a possibility that this function is differentiable.

Before setting out to do an investigation of the differentiability, let us take the liberty to do a few visual checks based on the calculations that we have performed until now. Maybe these figures can make us doubtful about the differentiability. Because we do not rely on purely visual proofs, we will then continue our reasoning as if we did not perform these visual checks.

3D view of function and tangent space. — We see here a three dimensional figure of the graph of the function $\frac{\partial f}{\partial x} (0,0) x + \frac{\partial f}{\partial y} (0,0) y$ , which is graphically the candidate tangent plane and the function $f (x, y)$ . We have doubts about the differentiability. We will have to follow our calculations.

We are going to define the quotient for this function

We define the quotient function, notation $q (h, k)$ in $(0,0)$ as

$q (h, k) = \frac{f (h, k) - f (0,0) - (\frac{\partial f}{\partial x} (0,0) h + \frac{\partial f}{\partial y} (0,0) k)}{| (h, k) |} .$

Remark that this quotient is the two variable equivalent of the one variable quotient

$q (h) = \frac{f (h) - f (0) - f^{'} (0) h}{h} .$

So please do not confuse this with $\frac{f (h) - f (0)}{h}$ , which is commonly called the differential quotient! To avoid any misunderstandings we call our $q$ from now on the quotient, notation $q (h, k)$ and not the differential quotient.

If $\lim_{(h, k) \to (0,0)} q (h, k) = 0,$

then the function $f (x, y)$ is by definition differentiable in $(0,0)$ . So we have to prove that the function

$\begin{matrix} q (h, k) = {\begin{matrix} \frac{h k (h^{2} - k^{2})}{{(h^{2} + k^{2})}^{3 ⁄ 2}} & if (h, k) \neq (0,0); \\ 0 & if (h, k) = (0,0) \end{matrix} \end{matrix}$

is continuous in $(0,0)$ .

Discussion of the continuity of q(h,k) in (0,0)

We investigate continuity with an $ϵ$ - $δ$ approach.

We take an arbitrary $ϵ > 0$ and we have to prove that $| q (h, k) - 0 | < ϵ$ . The problem is now to find a $δ > 0$ such that if $| (h, k) - (0,0) | < δ$ it follows that $| q (h, k) - 0 | < ϵ$ is valid.

When applying our function definition, we have then the following statements. Try to find a $δ$ such that if $| (h, k) - (0,0) | = \sqrt{h^{2} + k^{2}} < δ$ , we have that

$| \frac{h k (h^{2} - k^{2})}{{(h^{2} + k^{2})}^{3 ⁄ 2}} - 0 | < ϵ .$

We are looking for a function $ℓ (h, k)$ that is larger then or equal to the left hand side of this last inequality. This function $ℓ (h, k)$ has to have the property that it can be made smaller then $ϵ$ by carefully manipulating the value $δ .$ This is sufficient for our continuity proof.

$\begin{aligned} | \frac{h k (h^{2} - k^{2})}{{(h^{2} + k^{2})}^{3 ⁄ 2}} | & \leq \frac{\sqrt{h^{2} + k^{2}} \sqrt{h^{2} + k^{2}} ({\sqrt{h^{2} + k^{2}}}^{2} + {\sqrt{h^{2} + k^{2}}}^{2})}{{\sqrt{h^{2} + k^{2}}}^{3}} \\ \leq \frac{2 {\sqrt{h^{2} + k^{2}}}^{4}}{{\sqrt{h^{2} + k^{2}}}^{3}} \\ \leq 2 \sqrt{h^{2} + k^{2}} . \end{aligned}$

It is sufficient to take $δ = ϵ ⁄ 2$ . We can find a $δ$ , so we conclude that the function $q (h, k)$ is continuous. The function $f$ is differentiable.

3D view of the quotient. — We see here a three dimensional figure of the graph of the function $q (h, k)$ . This looks like a continuous function.

View of the contourplot of the quotient. — We see here a figure of the contour plot of the function $q (h, k)$ . Only level curves of level around $0$ come close to $(0,0)$ .

5. Continuity of the partial derivatives

Discussion of the continuity of the partial derivatives

We are looking for an alternative proof for the differentiability.

If both the partial derivative derivatives are continuous, then we have an alternative proof of the existence of the derivative. The condition that both of the partial derivatives are continuous is in fact too strong in the sense that it is not equivalent with differentiability only. But this criterion is in fact used by many instructors and textbooks, so it is interesting to take a look at it.

We know that the partial derivative to $x$ exists and is equal to

$\begin{matrix} \frac{\partial f}{\partial x} (x, y) = {\begin{matrix} \frac{y (x^{4} + 4 x^{2} y^{2} - y^{4})}{{(x^{2} + y^{2})}^{2}} & if (x, y) \neq (0,0), \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

We want to see if it is continuous or not.

Discussion of the continuity of the partial derivative of f(x,y) to x

We investigate continuity with an $ϵ$ - $δ$ approach.

We take an arbitrary $ϵ > 0$ and we have to prove that the inequality $| \frac{\partial f}{\partial x} (x, y) - \frac{\partial f}{\partial x} (0,0) | < ϵ$ holds under certain conditions. The problem is now to find a $δ > 0$ such that if $| (x, y) - (0,0) | < δ$ it follows that $| \frac{\partial f}{\partial x} (x, y) - \frac{\partial f}{\partial x} (0,0) | < ϵ$ is valid.

When applying our function definition, we have then the following statements. Try to find a $δ$ such that if $| (x, y) - (0,0) | = \sqrt{x^{2} + y^{2}} < δ$ , we have that

$| \frac{\partial f}{\partial x} (x, y) - \frac{\partial f}{\partial x} (0,0) | < ϵ .$

$\begin{aligned} | \frac{y (x^{4} + 4 x^{2} y^{2} - y^{4})}{{(x^{2} + y^{2})}^{2}} | \\ \leq \frac{| y | (x^{4} + 4 x^{2} y^{2} + y^{4})}{(x^{2} + y^{2})^{2}} \\ \leq \frac{\sqrt{x^{2} + y^{2}} ({\sqrt{x^{2} + y^{2}}}^{4} + 4 {\sqrt{x^{2} + y^{2}}}^{4} + {\sqrt{x^{2} + y^{2}}}^{4})}{{\sqrt{x^{2} + y^{2}}}^{4}} \\ \leq \frac{6 {\sqrt{x^{2} + y^{2}}}^{5}}{{\sqrt{x^{2} + y^{2}}}^{4}} \\ \leq 6 \sqrt{x^{2} + y^{2}} . \end{aligned}$

It is sufficient to take $δ = ϵ ⁄ 6$ . We can find a $δ$ , so we conclude that the function is continuous.

3dim view of the first partial derivative to x. — We see here a three dimensional figure of the graph of the first partial derivative $\frac{\partial f}{\partial x} (x, y)$ . This looks like a continuous function.

Contourplot of the derivative to x. — We see here a figure of the contour plot of the $\frac{\partial f}{\partial x} (x, y)$ . Only level curves of level around $0$ come close to $(0,0)$ .

Discussion of the continuity of the partial derivative of f(x,y) to y

By the symmetries in the function definition, we could argue on the basis of geometric arguments that the partial derivative to $y$ is also continuous. But we will give again a proof based on calculations.

We know that the partial derivative to $y$ exists and is equal to

$\begin{matrix} \frac{\partial f}{\partial y} (x, y) = {\begin{matrix} \frac{x^{5} - 4 x^{3} y^{2} - x y^{4}}{{(x^{2} + y^{2})}^{2}} & if (x, y) \neq (0,0), \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

We want to see if it is continuous or not.

Discussion of the continuity in (0,0)

We investigate continuity with an $ϵ$ - $δ$ approach.

We take an arbitrary $ϵ > 0$ and we have to prove the inequality $| \frac{\partial f}{\partial y} (x, y) - \frac{\partial f}{\partial y} (0,0) | < ϵ$ . The problem is now to find a $δ > 0$ such that if the inequality $| (x, y) - (0,0) | < δ$ holds, then it follows that $| \frac{\partial f}{\partial y} (x, y) - \frac{\partial f}{\partial y} (0,0) | < ϵ$ is valid.

When applying our function definition, we have then the following statements. Try to find a $δ$ such that if $| (x, y) - (0,0) | = \sqrt{x^{2} + y^{2}} < δ$ , we have that

$| \frac{\partial f}{\partial y} (x, y) - \frac{\partial f}{\partial y} (0,0) | < ϵ .$

$\begin{aligned} | \frac{\partial f}{\partial y} | & \leq | \frac{x^{5} - 4 x^{3} y^{2} - x y^{4}}{{(x^{2} + y^{2})}^{2}} | \\ \leq \frac{| x |^{5} + 4 | x |^{3} y^{2} + | x | y^{4}}{{(x^{2} + y^{2})}^{2}} \\ \leq \frac{{\sqrt{x^{2} + y^{2}}}^{5} + 4 {\sqrt{x^{2} + y^{2}}}^{5} + {\sqrt{x^{2} + y^{2}}}^{5}}{{\sqrt{x^{2} + y^{2}}}^{4}} \\ \leq \frac{6 {\sqrt{x^{2} + y^{2}}}^{5}}{{\sqrt{x^{2} + y^{2}}}^{4}} \\ \leq 6 \sqrt{x^{2} + y^{2}} . \end{aligned}$

It is sufficient to take $δ = ϵ ⁄ 6$ . We can find a $δ$ , so we conclude that the function is continuous.

6. Overview and summary

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x y (x^{2} - y^{2})}{x^{2} + y^{2}} & if (x, y) \neq (0,0), \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

continuity	yes
partial derivatives exist	yes
all directional derivatives exist	yes
differentiable	yes
partials are continuous	yes

7. One step further

Possible further investigations

We want to know if this function is further uneventful from the point of view of differentiability. Let us take a look at the second order partial derivative

\frac{\partial^{2} f}{\partial x^{2}} (x, y) = - \frac{4 x y^{3} (x^{2} - 3 y^{2})}{{(x^{2} + y^{2})}^{3}} .

Let us take a look of a three dimensional plot of this partial derivative to $y$ of the function.

Second order derivative twice to x gives discontinuity. — We see here a figure of the second order partial derivative $\frac{\partial^{2} f}{\partial x^{2}} (x, y)$ . It seems quite improbable that this second order partial derivative is continuous. We stop however our investigations here and leave this to the initiative of the interested reader.

3. List of exercises

We collected all exercises in the following table. The reader can read the solution in the text that can be downloaded.

Exercise 1.

$\begin{matrix} f (x, y) = {\begin{matrix} (x^{2} + y^{2}) \sin (x^{2} + y^{2}) & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 2.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x^{3} y}{x^{4} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 3.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{3 x^{2} y - y^{3}}{x^{2} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 4.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{y^{3} - x^{8} y}{x^{6} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 5.

$\begin{matrix} f (x, y) = {\begin{matrix} x^{3} \sin (\frac{1}{x^{2}}) + y^{3} \sin (\frac{1}{y^{2}}) & if x y \neq 0; \\ 0 & if x y = 0 . \end{matrix} \end{matrix}$

Exercise 6.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x | y |}{x^{2} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 7.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{y | x |^{3 ⁄ 2}}{x^{3} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 8.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{\sin (| x y | + x^{2})}{x} & if x \neq 0; \\ 0 & if x = 0 . \end{matrix} \end{matrix}$

Exercise 9.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x^{2} y^{2}}{x^{2} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 10.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x y^{4}}{x^{2} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 11.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x y^{2}}{x^{2} + y^{4}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 12.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{y {(x^{2} - y)}^{2}}{x^{6}} & if x \neq 0; \\ 0 & if x = 0 . \end{matrix} \end{matrix}$

Exercise 13

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x^{3} y - x y^{3}}{x^{2} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 14.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x^{2} y}{x^{6} + y^{2}} & if (x, y) \neq (0, 0); \\ 0 & if (x, y) = (0, 0) . \end{matrix} \end{matrix}$

Exercise 15.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x^{2} | y |^{5 ⁄ 4}}{x^{4} + y^{2}} & if (x, y) \neq (0, 0); \\ 0 & if (x, y) = (0, 0) . \end{matrix} \end{matrix}$

Exercise 16.

$\begin{matrix} f (x, y) = \sqrt[3]{x^{2} y} . \end{matrix}$

Exercise 17.

$\begin{matrix} f (x, y) = {\begin{matrix} x^{2} y \sin (\frac{1}{x}) & if x \neq 0; \\ 0 & if x = 0 . \end{matrix} \end{matrix}$

Exercise 18.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x y}{\sqrt{x^{2} + y^{2}}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 19.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x^{2} y}{x^{2} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 20.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x^{3} + y^{3}}{x^{2} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 21.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x^{3} y}{x^{6} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 22.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x^{5} y}{x^{8} + y^{4}} & if (x, y) \neq (0, 0); \\ 0 & if (x, y) = (0, 0) . \end{matrix} \end{matrix}$

Exercise 23.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x^{5} + y^{4}}{{(x^{2} + y^{2})}^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 24.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x y^{2}}{\sqrt{x^{2} + y^{2}} (x^{2} + y^{4})} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 25.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x y (x^{2} - y^{2})}{x^{2} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 26.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{y \sin (x y)}{x^{2} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 27.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{y^{3}}{x^{2} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 28.

$\begin{matrix} f (x, y) = {\begin{matrix} (x^{2} + y^{2}) \log (x^{2} + y^{2}) & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 29.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x^{2} + y^{2}}{x^{2} + y^{4}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 30.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x^{2} y^{2}}{x^{2} y^{2} + (x - y)^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 31.

$\begin{matrix} f (x, y) = {\begin{matrix} x^{2} (\sin (\frac{1}{x + y}) + 2) & if x + y \neq 0; \\ 0 & if x + y = 0 . \end{matrix} \end{matrix}$

Exercise 32.

$\begin{matrix} f (x, y) = {\begin{matrix} x \sin (\frac{1}{x}) + y \sin (\frac{1}{y}) & if x y \neq 0; \\ 0 & if x = 0 or y = 0 . \end{matrix} \end{matrix}$

Exercise 33.

$\begin{matrix} f (x, y) = {\begin{matrix} x^{2} \sin^{2} (\frac{y}{x}) & if x \neq 0; \\ 0 & if x = 0 . \end{matrix} \end{matrix}$

Exercise 34.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{\sin^{2} (x + y)}{| x | + | y |} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 35.

$\begin{matrix} f (x, y) = | x y | . \end{matrix}$

Exercise 36.

$\begin{matrix} f (x, y) = {\begin{matrix} y \sin (\frac{x}{\sqrt{| y |}}) & if y \neq 0; \\ 0 & if y = 0 . \end{matrix} \end{matrix}$

Exercise 37.

$\begin{matrix} f (x, y) = \sin (y) sgn (\sin (x)) . \end{matrix}$

Exercise 38.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{\sin (4 x \sqrt{| y |})}{\sqrt{| x y |}} & if x y \neq 0; \\ 0 & if x y = 0 . \end{matrix} \end{matrix}$

Exercise 39.

$f (x, y) = \max {x, y} .$

Exercise 40.

$\begin{matrix} f (x, y) = y \sqrt{| x |} . \end{matrix}$

Exercise 41.

$f (x, y) = {\begin{matrix} x \sin (\frac{1}{x^{2} + y^{2}}) & if (x, y) \neq (0, 0); \\ 0 & otherwise \end{matrix}$

Exercise 42.

$f (x, y) = \max {| x |, | y |} .$

Exercise 43.

$\begin{matrix} f (x, y) = {\begin{matrix} (x^{2} + y^{2}) \sin (\frac{1}{x + y}) & if x + y \neq 0; \\ 0 & if x + y = 0 . \end{matrix} \end{matrix}$

Exercise 44.

$\begin{matrix} f (x, y) = {\begin{matrix} (x^{2} + y^{2}) \sin (\frac{1}{x^{4} + y^{4}}) & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 45.

$\begin{matrix} f (x, y) = {\begin{matrix} (x^{2} + y^{2}) \sin (\frac{1}{x^{2} + y^{2}}) & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 46.

$\begin{matrix} f (x, y) = {\begin{matrix} x y^{2} \sin (\frac{1}{y}) & if y \neq 0; \\ 0 & if y = 0 . \end{matrix} \end{matrix}$

Exercise 47.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{| y | 2^{- \frac{| y |}{x^{2}}}}{x^{2}} & if x \neq 0; \\ 0 & if x = 0 . \end{matrix} \end{matrix}$

Exercise 48.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x y}{| x |} + x \sin (\frac{1}{y}) & if x y \neq 0; \\ 0 & if x = 0 or y = 0 . \end{matrix} \end{matrix}$

Exercise 49.

$\begin{matrix} f (x, y) = {\begin{matrix} x^{3} \sin (\frac{1}{x^{2}}) + y^{3} \sin (\frac{1}{y^{2}}) & if x y \neq 0; \\ 0 & if x y = 0 . \end{matrix} \end{matrix}$

Exercise 50.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{y \sqrt{x^{2} + y^{2}}}{| y |} & if y \neq 0; \\ 0 & if y = 0 . \end{matrix} \end{matrix}$

Exercise 51.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x^{2} y \sqrt{| y |}}{x^{4} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 52.

$\begin{matrix} f (x, y) = \sqrt[3]{x} \sqrt[3]{y} . \end{matrix}$

Exercise 53.

$\begin{matrix} f (x, y) = {\begin{matrix} \sin (\frac{x^{2} + y^{2}}{x^{4} + y^{4}}) & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 54.

$\begin{matrix} f (x, y) = \sqrt{\sin^{2} (x) + \sin^{2} (y)} . \end{matrix}$

Exercise 55.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{| x | + | y | - | x + y |}{{(x^{2} + y^{2})}^{1 ⁄ 5}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 56.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{| x | + | y | - | x + y |}{{(x^{2} + y^{2})}^{1 ⁄ 2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 57.

$\begin{matrix} f (x, y) = {\begin{matrix} \sin (\frac{1}{x^{2} + y^{2}}) & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 58.

$See below.$

Exercise 59.

$\begin{matrix} f (x, y) = \min {x, y} = \frac{x + y}{2} - \frac{| x - y |}{2} . \end{matrix}$

Exercise 60.

$\begin{matrix} f (x, y) = | x | + | y | . \end{matrix}$

Exercise 61.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x y}{x^{2} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 62.

$\begin{matrix} f (x, y) = {(\sqrt[3]{x} + \sqrt[3]{y})}^{3} . \end{matrix}$

Exercise 63.

$f (x, y) = \sqrt{\sin^{2} (x) + \sin^{2} (y)} .$

Exercise 64.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{(x + y)^{2}}{x^{2} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 65.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{\log (x^{2} y^{2})}{x^{2} + y^{2}} & if x y \neq 0; \\ 0 & if x y = 0 . \end{matrix} \end{matrix}$

Exercise 66.

$See below.$

Exercise 67.

$\begin{matrix} f (x, y) = {\begin{matrix} x + y & if x y \neq 0; \\ 0 & if x = 0 or y = 0 . \end{matrix} \end{matrix}$

Exercise 68.

$\begin{matrix} f (x, y) = {\begin{matrix} x^{2} (1 - \cos (\frac{y}{x})) & if x \neq 0; \\ 0 & if x = 0 . \end{matrix} \end{matrix}$

Exercise 69.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{\sin (x - y)}{\sqrt{| x |} + \sqrt{| y |}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 70.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{x y - \sin (x) \sin (y)}{x^{2} + y^{2}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 71.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{\log (| x | + e^{| y |})}{\sqrt{x^{2} + y^{2}}} & if (x, y) \neq (0,0); \\ 0 & if (x, y) = (0,0) . \end{matrix} \end{matrix}$

Exercise 72.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{(x - y) y e^{- \frac{1}{x^{2}}}}{(x - y)^{2} + 2 e^{- \frac{2}{x^{2}}}} & if x \neq 0; \\ 0 & if x = 0 . \end{matrix} \end{matrix}$

Exercise 73.

$\begin{matrix} f (x, y) = \sqrt{4 - x^{2} - y^{2}} . \end{matrix}$

Exercise 74.

$See below.$

Exercise 75.

$See below.$

Exercise 76.

$\begin{matrix} f (x, y) = ⌊ x + y ⌋ . \end{matrix}$

Exercise 77.

$\begin{matrix} f (x, y) = {\begin{matrix} \frac{\cos (x y) - 1}{x^{2} y^{2}} & if x y \neq 0; \\ \frac{- 1}{2} & if x y = 0 . \end{matrix} \end{matrix}$

Exercise 78.

$See below.$

The exercises marked in the table with “See below” are listed here.

Exercise 58.

$\begin{matrix} f (x, y) = {\begin{matrix} x \sin (\frac{1}{x}) & if y = 0 and x \neq 0; \\ y \sin (\frac{1}{y}) & if x = 0 and y \neq 0; \\ 0 & elsewhere. \end{matrix} \end{matrix}$

Exercise 66.

$\begin{matrix} f (x, y) = {\begin{matrix} x^{2} \sin (\frac{1}{x}) + y^{2} \sin (\frac{1}{y}) & if x y \neq 0; \\ x^{2} \sin (\frac{1}{x}) & if x \neq 0 and y = 0; \\ y^{2} \sin (\frac{1}{y}) & if y \neq 0 and x = 0; \\ 0 & if x = 0 and y = 0 . \end{matrix} \end{matrix}$

Exercise 74.

$\begin{matrix} f (x, y) = {\begin{matrix} 1 & if x^{2} = y and x \neq 0; \\ 0 & otherwise. \end{matrix} \end{matrix}$

Exercise 75.

$\begin{matrix} f (x, y) = {\begin{matrix} (x + 1)^{2} + (y + 1)^{2} - 2 & if x < y and y < 2 x and x > 0; \\ 0 & elsewhere. \end{matrix} \end{matrix}$

Exercise 78.

$\begin{matrix} f (x, y) = {\begin{matrix} x^{2} + y^{2} & if x < y and y < 2 x and x > 0; \\ 0 & elsewhere. \end{matrix} \end{matrix}$

4. Download the exercises text document

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Exercises on the differentiability of functions in two variables

1. Overview of differentiability of functions of two variables

1. Examples of exercises

Two completely solved exercises

1. Continuity of the function

2. Partial derivatives

Discussion of the partial derivative to x in f(x, y)

Discussion of the partial derivative to y in f(x, y)

3. Directional derivatives

4. Differentiability

Some preliminary visual tests

Defining the quotient function

Discussion of the continuity of q(h,k) in (0,0)

5. Discussion of the continuity of the partial derivatives

6. Overview and summary

7. One step further

Possible further investigations

7. Alternative proof for the non differentiability

8. Behaviour of the gradient

1. Continuity of the function

2. Partial derivatives

Discussion of the partial derivative of f(x,y) in x

Discussion of the partial derivative of f(x,y) in x

3. Directional derivatives

Discussion of the directional derivatives in (0,0)

4. Differentiability

Some preliminary visual tests

We are going to define the quotient for this function

Discussion of the continuity of q(h,k) in (0,0)

5. Continuity of the partial derivatives

Discussion of the continuity of the partial derivatives

Discussion of the continuity of the partial derivative of f(x,y) to x

Discussion of the continuity of the partial derivative of f(x,y) to y

Discussion of the continuity in (0,0)

6. Overview and summary

7. One step further

Possible further investigations

3. List of exercises

4. Download the exercises text document

5. Searching for further information.