# Math Help - partial derivatives

1. ## partial derivatives

1. If it's given that A: R2 -> R is defined by A(x, y)={1, if xy greater than or equal to 0.
and 0, if xy < 0.
(a) how do i show that the partial derivatives Ax(0) and Ay(0) exist? (Note that Ax and Ay doesn't mean A*x and A*y. I just mean the first order derivatives.)
(b) How do I prove that A is not continuous at 0?

2. If A(x, y, z) = xzsin2y + ye^z
(a) How do I find all the mixed partial derivatives of A and verify Clairant's Theorem holds?
(b) Will Ayxz = Ayzx? Will I know this without actually calculating?
(c) Will Axyy=Ayzx?
(d) Will Ayzy=Azyy?

2. Originally Posted by Taurus3
1. If it's given that A: R2 -> R is defined by A(x, y)={1, if xy greater than or equal to 0. and 0, if xy < 0. (a) how do i show that the partial derivatives Ax(0) and Ay(0) exist? (Note that Ax and Ay doesn't mean A*x and A*y. I just mean the first order derivatives.)

For all neighbourhood of $(0,0)$ the function is not elemental. So, apply just the definition of partial derivative.

3. Originally Posted by Taurus3
2. If A(x, y, z) = xzsin2y + ye^z
(a) How do I find all the mixed partial derivatives of A and verify Clairant's Theorem holds?

What particular difficulties have you had?

4. I feel really bad saying this. But I just don't get it. Same with the second question. Can you show the work for me? I know I'm asking a bit too much.

5. $\frac{\partial f}{\partial x}(x_0, y_0)= \lim_{h\to 0}\frac{f(x_0+ h, y_0)- f(x_0, y_0)}{h}$
use that definition to find the limit at x= 0, y= 0.

Since f(x,0)= 1 for all x, that calculation is pretty close to being trivial.

6. wait, so that's the final answer? It proves it?

7. I'm having a bit of difficulty reading question 1.

Are you asking to show that

$A_x(x,0)$ and $A_y(0,y)$ exist?

For 1(b), "not continuous at 0"? I assume you mean the origin, the point (0,0)? There is no just "0" in $\mathbb{R}^2$.

To show $A(x,y)$ is not continuous at $(0,0)$, you need to show

$\lim_{(x,y)\rightarrow(0,0)}A(x,y)\neq A(0,0)=1$. (In particular, this is true if the limit happens to not even exist, for example.)

So let's show $\lim_{(x,y)\rightarrow(0,0)}A(x,y)\neq 1$. Recall that, with functions of one variable, there are only two ways to approach an $x$-value: from the left or right. In $\mathbb{R}^2$, however, you can approach $(0,0)$ in infinitely many different ways; for example, along the $x$-axis, the $y$-axis, the parabola $y=x^2$, etc, etc.

In this case, let's approach $(0,0)$ along the line $y=-x$ (note that the point $(0,0)$ is actually on this line, otherwise this would make no sense). So we are computing $\lim_{(x,y)\rightarrow(0,0)}A(x,y)$ along the line $y=-x$; that is, the only points we are concerning ourselves with are those for which $y=-x$. Thus the limit becomes

$\lim_{(x,y)\rightarrow(0,0)}A(x,y)=\lim_{(x,y)\rig htarrow(0,0)}A(x,-x)$.

But what is $A(x,-x)$. We can assume $x\neq 0$ because, on the line we are on, that would put us at the origin. But we can't be AT the origin if we are trying to take the limit as we APPROACH the origin. You can show (it's very, very easy...) that, whenever $x\neq 0,x\cdot (-x)<0$. This means $A(x,-x)=0$. Therefore

$\lim_{(x,y)\rightarrow(0,0)}A(x,y)=\lim_{(x,y)\rig htarrow(0,0)}A(x,-x)=\lim_{(x,y)\rightarrow(0,0)}(0)=0$.

Now, a limit exists if and only if it is the same along any path whatsoever. Since the limit along this path is 0, if the limit were to actually exist, it would have to be equal to 0. But $A(0,0)=1$, which means we would have

$\lim_{(x,y)\rightarrow(0,0)}A(x,y)=0\neq A(0,0)=1$.

So the function cannot be continuous at the origin.

(NOTE: I'm not saying the limit IS 0; just that, by our above computation, it would have to be 0 IF IT EXISTED. You can actually show that the limit for this function does not exist at all.)

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Isn't question 2 straightforward? It's just asking you to compute partial derivatives.