AP Calc BC question

What's the gradient of the following?

5xyz + 3yz + 2z

http://mathworld.wolfram.com/Gradient.html

(partial x, partial y, partial z)

>>3

<partial x, partial y, partial z>

[partial x, partial y, partial z]

{partial x, partial y, partial z}

>>6
no.

Can a vector not be considered a set?

>>8
Sets aren't ordered.

>>9
ordered sets are ordered.

>>10
ordered n-tuples and ordered sets with n elements are different things.

>>11
Also if you need another reason why vectors aren't sets, sets demand uniqueness.

In ZF isn't every element of a set itself set?  So in this sense vectors are sets, just don't think that their components are their elements.

>>11
What is the difference then?

>>12
Uniqueness?  In what sense?  For any vector, there are no other vectors equal to it, just like a set.

>>14
An ordered set is a set and a relation on that set satisfying certain properties. An ordered n-tuple is a list (colloquially) of n items in which order matters. If we take the set {a,b,c} with some order <=, then {b,a,c} with the same order is the same ordered set. If we take the ordered n-tuple (a,b,c), (b,a,c) is a different ordered n-tuple (except in the case where a = b).

As for uniqueness, the elements of a set have to be unique. Even if we ignore the ordering problem for a moment, you can't represent vectors as sets (in any nice way). The representation of (0,0,1) is {0,0,1}, which (by uniqueness) is equivalent to {0,1}. The representation of (0,1,1) is {0,1,1}, which is equivalent to {0,1}. Obviously that's problematic.

>>15
Thanks, I forgot that ordered sets must also have unique elements.

   partial x |
< partial y |
   partial z |

grad(5xyz + 3yz + 2z) = (5yz, 5xz + 3z, 5xy + 3y + 2)

>>18
Why the fuck did you go and do that?

the gradient is a linear function T satisfying:
lim_{h -> 0} (f(a + h) - f(a) - T(a))/|h| = 0

I feel the need, the need for weed!

Marijuana MUST be legalized.