Conflicting result in derivative of composite function

[toforfiltum]

Homework Statement

Let $$f(x,y)=\begin{cases} \frac{x^2y}{x^2+y^2} \space & \text{if} \space(x,y)\neq(0,0)\\0 \space & \text{if} \space(x,y)=(0,0)\end{cases}$$

a) Use the definition of the partial derivative to find $f_x(0,0)$ and $f_y(0,0)$.

b) Let a be a nonzero constant and let $x(t)=(t,at)$. Show that $f\circ x$ is differentiable, and find $D(f\circ x)(0)$ directly.

c) Calculate ##Df(0,0)Dx(0). How can you reconcile your answer in part b) and the chain rule?

The Attempt at a Solution

a) By going to the limit definition $f_x(0,0)=0$ and $f_y(0,0)=0$.

b) $$f \circ x(t,at)=\frac{t^2(at)}{t^2(1+a^2)}$$
$$=\frac {at}{1+a^2}$$

Hence,$$D(f \circ x)=\frac{a}{1+a^2}$$ for all values of t.

c) According to answer in (a), $$Df(0,0)=(0,0)$$
$$Dx(0)= \left(\frac 1 a\right)$$
The product of both matrices yields zero.

I think that this conflicting result shows that at $(0,0)$, there is no differentiability. Maybe along the line $x(t)$, there is a derivative, but not for other paths going to $(0,0)$.

I don't know, I'm just guessing here. Can someone help me?

Thanks.

 

[Ray Vickson]

Homework Statement

Let $$f(x,y)=\begin{cases} \frac{x^2y}{x^2+y^2} \space & \text{if} \space(x,y)\neq(0,0)\\0 \space & \text{if} \space(x,y)=(0,0)\end{cases}$$

a) Use the definition of the partial derivative to find $f_x(0,0)$ and $f_y(0,0)$.

b) Let a be a nonzero constant and let $x(t)=(t,at)$. Show that $f\circ x$ is differentiable, and find $D(f\circ x)(0)$ directly.

c) Calculate ##Df(0,0)Dx(0). How can you reconcile your answer in part b) and the chain rule?

The Attempt at a Solution

a) By going to the limit definition $f_x(0,0)=0$ and $f_y(0,0)=0$.

b) $$f \circ x(t,at)=\frac{t^2(at)}{t^2(1+a^2)}$$
$$=\frac {at}{1+a^2}$$

Hence,$$D(f \circ x)=\frac{a}{1+a^2}$$ for all values of t.

c) According to answer in (a), $$Df(0,0)=(0,0)$$
$$Dx(0)= \left(\frac 1 a\right)$$
The product of both matrices yields zero.

I think that this conflicting result shows that at $(0,0)$, there is no differentiability. Maybe along the line $x(t)$, there is a derivative, but not for other paths going to $(0,0)$.

I don't know, I'm just guessing here. Can someone help me?

Thanks.

Don't guess; verify.

You now know the definition of differentiability at $(x,y) = (0,0)$, and can check whether or not your $f$ verifies that definition. If you cannot check it in cartesian coordinates, try switching to polar coordinates.

 

[toforfiltum]

Don't guess; verify.

You now know the definition of differentiability at $(x,y) = (0,0)$, and can check whether or not your $f$ verifies that definition. If you cannot check it in cartesian coordinates, try switching to polar coordinates.

Ah, switching to polar coordinates tells me that $f$ is indeed differentiable at $(0,0)$, because the limit exists and is equal to zero. This invalidates my argument above.

Does this mean that $x$ may not be differentiable at $(0,0)$?

 

[toforfiltum]

@Ray Vickson Yes, what I say may be true. At $x(0)$, the limits of the 2 functions are not equal to each other.

For $t$, $$\lim_{h \to 0}\frac{x(h)-x(0)}{h}$$
$$=\frac{h-0}{h}=1$$

For $at$, $$\lim_{h \to 0}\frac{x(h)-x(0)}{h}$$
$$=\frac{ah-0}{h}=a$$

Since the limits are not equal to each other, $x$ is not differentiable at 0.

Hence chain rule cannot be applied here.

Is this right?

 

[toforfiltum]

I realized I have mixed up between continuity and differentiability. In post #3, I should have said continuous, not differentiable.

$D(f \circ x)(0)$ works because the input of $f(x,y)$ now becomes $f(t,at)$, and $x(t)$ is defined for all values of $t$. So maybe this composition helps to sidestep the condition where $f(x,y)=0$ if $(x,y)=(0,0)$.

According to my TA,the chain rule doesn't work here because $f$ is not differentiable at $(0,0)$ since it is undefined at $(0,0)$.

And I think I can prove this.

$$\lim_{(x,y) \to (0,0)} \frac{f(x,y) -0}{\|(x,y)-(0,0)\|}$$
$$=\frac{x^2y}{(x^2+y^2)^\frac{3}{2}}$$
When function is approached along $y=0$ and $x=0$, limit is zero.

But when function is approached along $y=mx$ as $x \to 0$, a non zero value is obtained. Hence function is not differentiable at $(0,0)$.

This explains why the chain rule does not work, but it still baffles me why the differentiation of the composite function works.

 

[Ray Vickson]

I realized I have mixed up between continuity and differentiability. In post #3, I should have said continuous, not differentiable.

$D(f \circ x)(0)$ works because the input of $f(x,y)$ now becomes $f(t,at)$, and $x(t)$ is defined for all values of $t$. So maybe this composition helps to sidestep the condition where $f(x,y)=0$ if $(x,y)=(0,0)$.

According to my TA,the chain rule doesn't work here because $f$ is not differentiable at $(0,0)$ since it is undefined at $(0,0)$.

And I think I can prove this.

$$\lim_{(x,y) \to (0,0)} \frac{f(x,y) -0}{\|(x,y)-(0,0)\|}$$
$$=\frac{x^2y}{(x^2+y^2)^\frac{3}{2}}$$
When function is approached along $y=0$ and $x=0$, limit is zero.

But when function is approached along $y=mx$ as $x \to 0$, a non zero value is obtained. Hence function is not differentiable at $(0,0)$.

This explains why the chain rule does not work, but it still baffles me why the differentiation of the composite function works.

Why would you say that the reason $f$ is not differentiable at $(0,0)$ is that it is undefined at $(0,0)$. That is false reasoning: $f$ is defined at the origin---just go back and look at the definition of $f$ you gave in Post #1. It is true that $f$ is not differentiable at the origin, but for a different reason. Can you tell us why? Look at the definition of differentiability, and see if your $f$ verifies that definition at the origin.

 

[toforfiltum]

Why would you say that the reason $f$ is not differentiable at $(0,0)$ is that it is undefined at $(0,0)$. That is false reasoning: $f$ is defined at the origin---just go back and look at the definition of $f$ you gave in Post #1. It is true that $f$ is not differentiable at the origin, but for a different reason. Can you tell us why? Look at the definition of differentiability, and see if your $f$ verifies that definition at the origin.

That's not what I said, my TA said it. That didn't make sense to me, because like you said, it is defined at that point.

And I did use the definition of differentiability in the post above to show that it is not differentiable at that point. Perhaps I will type everything out this time.

Since partial derivatives at $(0,0)$, are both zero and function is defined as zero at $(0,0)$,

$$\lim_{(x,y) \to (0,0)} \frac{f(x,y)-0}{\|(x,y)-(0,0)\|}$$
$$=\frac{x^2y}{(x^2+y^2)^\frac{3}{2}}$$

Along the line $y=0$, $\lim_{x \to 0}=0$

Along the line $x=0$, $\lim_{y \to 0}=0$

Along the line $y=mx$ as $x \to 0$, $$\lim_{y=mx \to 0} \frac{mx^3}{x^3(1+m^2)^\frac{3}{2}} \neq 0$$

Hence, function is not differentiable at $(0,0)$.

 

[Ray Vickson]

That's not what I said, my TA said it. That didn't make sense to me, because like you said, it is defined at that point.

And I did use the definition of differentiability in the post above to show that it is not differentiable at that point. Perhaps I will type everything out this time.

Since partial derivatives at $(0,0)$, are both zero and function is defined as zero at $(0,0)$,

$$\lim_{(x,y) \to (0,0)} \frac{f(x,y)-0}{\|(x,y)-(0,0)\|}$$
$$=\frac{x^2y}{(x^2+y^2)^\frac{3}{2}}$$

Along the line $y=0$, $\lim_{x \to 0}=0$

Along the line $x=0$, $\lim_{y \to 0}=0$

Along the line $y=mx$ as $x \to 0$, $$\lim_{y=mx \to 0} \frac{mx^3}{x^3(1+m^2)^\frac{3}{2}} \neq 0$$

Hence, function is not differentiable at $(0,0)$.

OK, that is good enough.

Another way is to ask: are there constants $a$ and $b$ such that $f(x,y) = f(0,0) + a x + by + e(x,y)$, where the "error" term $e(x,y)$ is of higher order in small $\|(x,y)\|$ (that is, where $e(x,y)/ \|(x,y)\| \to 0$ as $(x,y) \to (0,0)$)? If you had such constants $a,b$, the would, necessarily, have to equal $f_x(0,0)$ and $f_y(0,0)$, respectively, because $f(x,0) - f(0,0) = ax + e(x,0)$, so $[f(x,0)-f(0,0)]/x \to a$ as $x \to 0$, etc. However, we have seen already that $f_x = f_y = 0$ at the origin, so if $f$ were differentiable at the origin we would have to have $f(x,y) = 0 x + 0 y + e(x,y) = e(x,y)$ of "higher order"; that is, we would have to have $f(x,y)/\|(x,y)\| \to 0$ as $\|(x,y)\| \to 0$. Is that true? Well, using polar coordinates we have $f(x,y) = r \cos^2(\theta) \sin(\theta)$ and, of course, $\|(x,y)\| = r$, so $f(x,y)/\|(x,y)\| = \cos^2(\theta) \sin(\theta)$. For most $\theta$ this does not go to 0 as $r \to 0$.