Differentiability of a multivariable function?

[Chris L]

As a preface, this question is taken from Vector Calculus 4th Edition by Susan Jane Colley, section 2.3 exercises.

Homework Statement

"Explain why each of the functions given in Exercises 34-36 is differentiable at every point in its domain."

34. $$xy - 7x^8y^2 cosx$$
35. $$\frac{x + y + z}{x^2 + y^2 + z^2}$$
36. $$(\frac{xy^2}{x^2 + y^4}, \frac{x}{y} + \frac{y}{x})$$

Homework Equations

Not equations, rather theorems, but basically for a function to be differentiable at a point a the function itself must exist at a and each of its partial derivatives must be continuous at a.

The Attempt at a Solution

It is immediately clear that 34 must be differentiable everywhere, no objections to this question. However, I have problems with 35 and 36:

35: First and foremost, the function isn't defined at (0,0,0) (as 0/0 isn't defined). If you compute its partial derivatives, they also are not defined at (0,0,0) (also equal to 0/0), so from the definitions in the book, the function isn't differentiable everywhere. Am I misinterpreting the question/have the wrong conditions for differentiability or is the question in the book worded poorly (perhaps the intended problem is to explain whether or not each function given is differentiable everywhere?)

36: Similar to 35, this function isn't defined when either x or y is 0, and its partial derivatives aren't either.


Thanks in advance.

 

[MostlyHarmless]

0/0 is an indeterminate form. Like ∞/∞ or 0^0.

 

[Chris L]

From the math courses I've taken, I've always been told that indeterminate forms mean that the function isn't defined at that point, rather than defining the value at that point to be equal to the limit. I can see the advantages of that convention, but that doesn't account for 36, where for example if you choose x = 0 and y = 2, the second component is 2/0 (which isn't indeterminate) and thus that function still isn't differentiable everywhere?

 

[Vargo]

The question says to verify that the functions are differentiable in their domain. So you don't have to check points like (0,0) because they don't belong to the domain.

For example, 1/x is continuous and infinitely differentiable everywhere in its domain because 0 is not in its domain.

 

[MostlyHarmless]

Ah, I just read the instructions again. It says explain why it is differentiable at every point in its "domain" If the function is not defined at that point, then that point is not in the domain.

 

[Chris L]

Oh wow, I can't read it seems, that would explain it. Thanks for your help!