Question

Given \(g(x,y) = \sqrt[3]{xy}\). Show that \(g_x(0,0)\) and \(g_y(0,0)\) exist by supplying values for them

Answer 1

If I am doing it right, then
\[g_x= \frac{y}{3(xy)^\frac{2}{3}}\]
If I put (x,y) = (0,0) into \(g_x\), then I’ll go into trouble

Answer 2

This is a bizarre question.

Answer 3

It's clearly not defined at those points. Do they mean to say the limit exists.

Answer 4

Part (a) of the question asks about the continuity of g(x,y), so I don't think it is asking if the limit exists.

Answer 5

And for the part after this, it actually asks if \(g_x\) and \(g_y\) are continuous at (0,0).

Answer 6

Okay, here is the thing you have to understand: \(|x|\) is not differentiable at \(0\).
The same thing applies \(\sqrt[2n]{x}\)

Answer 7

Since \(\sqrt{}\) is the positive root.

Answer 8

Okay how about this...

Answer 9

\[
yy^{-2/3}=y^{1/3}
\]Does that help?

Answer 10

But you still get 0/0

Answer 11

You're not getting 0 out of that denominator as far as I can tell.

Answer 12

Why not?
\[g_x= \frac{y}{3(xy)^\frac{2}{3}} = \frac{y^\frac{1}{3}}{3x^\frac{2}{3}}\]

Answer 13

Try going back to the definition of a partial derivative maybe?

Answer 14

I don't know how to go about this, but it seems wrong.

Answer 15

Do you mean I have to try finding \(g_x\) by the definition?

Answer 16

The only thing I can think of is some loss of generalization that happened because or using differentiation rules.

Answer 17

Do you need a refresh on the definition?

Answer 18

\[
g_x(0,0) = \lim_{h\to 0}\frac{g(h,0)-g(0,0)}{h}
\]

Answer 19

That gives you \(0\)... woah.

Answer 20

I guess using differentiation rules CAN screw you over.

Answer 21

Wait...
\[g_x(0,0) = \lim_{h\to 0}\frac{g(h,0)-g(0,0)}{h} = \lim_{h\to 0}\frac{\sqrt[3]{h(0)}-\sqrt[3]{0\times 0}}{h} \]Right?

Answer 22

Yes.

Answer 23

What is the issue?

Answer 24

\[g_x(0,0) = \lim_{h\to 0}\frac{g(h,0)-g(0,0)}{h} = \lim_{h\to 0}\frac{0}{h}\]Hmm..

Answer 25

Did you have quote extension?

Answer 26

Okay. \(0/h=0\)

Answer 27

Mind = blown

Why applying differentiation rules doesn't work here? Or, where did I make a mistake?

Answer 28

I have a couple of suspicions.

Answer 29

Yes?

Answer 30

One is that the differentiation rules don't work the same way for partial derivatives.

Answer 31

But here's what Wolf gives us...
http://www.wolframalpha.com/input/?i=partial+derivative+%5Csqrt%5B3%5D%7Bxy%7D

Answer 32

The other is that there is an underlying assumption the differentiation rules are making which we are overlooking.

Answer 33

http://www.wolframalpha.com/input/?i=partial+derivative+%5Csqrt%5B3%5D%7Bxy%7D+at+%280%2C0%29

The Wolf is not always perfect.

Answer 34

I actually am trying to think of ways that this could happen with a single variable function, but I don't think it is possible.

Answer 35

At (0,0), it is not differentiable, what's the problem?

Answer 36

It is differentiable by the definition though...

Answer 37

This one gives the general case, and if xy=0
http://www.wolframalpha.com/input/?i=partial+derivative+%5Csqrt%5B3%5D%7Bxy%7D

This one gives a specific case, if x=y=0
http://www.wolframalpha.com/input/?i=partial+derivative+%5Csqrt%5B3%5D%7Bxy%7D+at+%280%2C0%29

I just don't understand how come is it differentiable, especially when you look at the graph..

Answer 38

* how come it is

Answer 39

http://www.proofwiki.org/wiki/Definition:Differentiable

Answer 40

Because our concept of differentiable for single variable functions doesn't quite apply to multi variable functions the same way.

Answer 41

You're only differentiating in a single direction.

Answer 42

Okay, just consider: \[
g(x,0)=0
\]

Answer 43

Clearly \(0\) is differentiable.

Answer 44

Are they really the same? Differentiate the function first, the evaluate it, and evaluate it first, then differentiate it.

Answer 45

That is why it works. When you project it onto the \(xz\) plane it becomes differentiable. I think this phenomenon is exclusive to multi-variable derivatives. That makes it a uncommon phenomenon.

Answer 46

@Callisto I'm not actually sure if they're the same by definition, but I think that the "projection" I'm talking about it part of the intent.

Answer 47

the intent of partial derivatives.

Answer 48

Alright... So, to play safe, it is better to start with the definition, right?

Answer 49

The definition is the true answer.

Answer 50

Differentiation rules are a short cut.

Answer 51

I think the insight to be taken from this is:

If a multi-variable function returns an indeterminate form at a point, then going back to the original definition could provide some insight.

Answer 52

That definition gives you the mathematical power of a limit. They are powerful devices.

Answer 53

Ok... Thanks for your help!!