Contradiction?

[Mao]

but why does its derivative function has to be absolutely true?

have you checked that this function complies with the rules of differentiability at x=0? [i'm pretty sure vertical tangents are exceptions to differentiation, and this is marked by its derivative's domain. this, however, does not imply it doesn't have a gradient]

if you differentiate this with vector calculus, i'm sure you'll end up with a unit tangent of

doing vector calculus now [Captain: dont laugh, haha]

the relationship definining this function is continuous, and can be described by the relationship:

where t is a parameter.

this relationship can then be defined as a space-curve relative to O as:





hence its unit tangent can be described as:





yay vertical tangent =]

[Fitness]

the relationship definining this function is continuous, and can be described by the relationship:

Incorrect.
if y³ = x
then y = x^1/3 and y = -x^1/3

you can't tell me that x^1/3 = -x^1/3

[Mao]

what I meant to say was, derivatives are not the be-all and end-all when it comes to gradients.

it is absurd to say that a unit circle does not have a gradient at x=1, yet this certainly cannot be found by differentiation.

differetiation is based on the change of one variable against another. one the independant variable does not change, then it fails. i.e. vertical tangents.

[Fitness]

what I meant to say was, derivatives are not the be-all and end-all when it comes to gradients.

it is absurd to say that a unit circle does not have a gradient at x=1, yet this certainly cannot be found by differentiation.

differetiation is based on the change of one variable against another. one the independant variable does not change, then it fails. i.e. vertical tangents.

An end point doesn't have a vertical tangent... It has no tangent at all. That's what I am trying to say.

[bigtick]

Infinity is not a number; it is the name for a concept.
Infinity is a well defined concept, but 1/0 is undefined.
1/0 not= oo, but lim(x->0) of 1/x = oo.
The gradient of a vertical tangent is not= 1/0, it is = oo.

[Mao]

the relationship definining this function is continuous, and can be described by the relationship:

Incorrect.
if y³ = x
then y = x^1/3 and y = -x^1/3

you can't tell me that x^1/3 = -x^1/3

absolutely absurd
,

what I meant to say was, derivatives are not the be-all and end-all when it comes to gradients.

it is absurd to say that a unit circle does not have a gradient at x=1, yet this certainly cannot be found by differentiation.

differetiation is based on the change of one variable against another. one the independant variable does not change, then it fails. i.e. vertical tangents.

An end point doesn't have a vertical tangent... It has no tangent at all. That's what I am trying to say.

but it doesnt have an end-point. the cubic root function is a continuous function defined for

I think you are confusing the cubic root function with the square root function.

[Fitness]

*commits an hero*
I realised just before I went to bed last night what I did :/

[Captain]

doing vector calculus now [Captain: dont laugh, haha]

I laughed hard.