Posts made by Nukchebi0

My period argues with your success.

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I like LL. It seems a perfect balance between strategy and luck.

They should. It should be made a house “official” rule for tournaments.

The rule stating that even if one unit is amphibiously invading and 100 marching overland, all are considered amphibiously invading and can’t retreat.

The rule about amphibious invasions is asinine. It should apply only to the units offloaded from the transports, not those moving over land as well.

It equals 6, not zero.

I need this for tomorrow.

It is x to the third times y, not x to the 3y.

Consider a curve of x(y^2)-x^(3)y=6.

The derivative is: (3(x^2)y-y^2) / (2xy-x^3)

I need to find when the derivative does not exist (tangent line vertical).

So far, as we only need to see if the denominator is 0, I set 2xy-x^3=0.

Then, I factored this to get x(2y-x^2)=0.

I know from here that when x=0, there is a vertical tangent line. However, I think there is more than that. What should I do to find the rest, or if I am doing this wrong, how do I solve it?

@Romulus:

@Nukchebi0:

@Romulus:

@Nukchebi0:

What is the derivative of x^(1/3) and -x(1/3)? I tried finding an equation that eliminated 0 from the denominator, but failed. I also tried using the “cheat” method (x^n ->
nx^(x-1), but couldn’t see how this was implemented.

If possible could you show the algebra steps? I don’t need the calculus notation, but would like to know how it works, not just the answer.

You said that you tried to eliminate the 0 from the denominator.
I answered that it is a problem of discontinuity. Deriving a function you are worsening the function properties. Example: x^1/3 is continuous and defined in all the se of real number R. Its derivated function (1/3) * x^1/3 is discontinuous in the point 0, where it is not defined.
If you need to do some operation that uses those two derivative in the point 0, you may not use them, becasue for x = 0 the two functions are not defined, their value simply does not exist.

I see now, thanks.

Excuse me, in the post you have quoted I wrote the derivate worng it is : (1/3)*x^(-2/3)!
But the reasoning is right.
I have corrected the original post!

I figured that you did. I make a ton of mistakes of that ilk.

@cystic:

boy - i hope that Romulus gets a lot of positive Karma for this . . . .

And yes, he will receive positive karma.

Edit: It seems I can’t. I try and I get a message saying server verification has failed. The suggested fix does nothing.

@Romulus:

@Nukchebi0:

What is the derivative of x^(1/3) and -x(1/3)? I tried finding an equation that eliminated 0 from the denominator, but failed. I also tried using the “cheat” method (x^n ->
nx^(x-1), but couldn’t see how this was implemented.

If possible could you show the algebra steps? I don’t need the calculus notation, but would like to know how it works, not just the answer.

You said that you tried to eliminate the 0 from the denominator.
I answered that it is a problem of discontinuity. Deriving a function you are worsening the function properties. Example: x^1/3 is continuous and defined in all the se of real number R. Its derivated function (1/3) * x^1/3 is discontinuous in the point 0, where it is not defined.
If you need to do some operation that uses those two derivative in the point 0, you may not use them, becasue for x = 0 the two functions are not defined, their value simply does not exist.

I see now, thanks.

@Romulus:

d/dx(x^n) = n x^n-1 is the right rule to use.

d/dx (x^1/3) = (1/3) * x^(1/3 - 1) = (1/3) *x^(-2/3)

d/dx (x^-1/3) = (-1/3) *  x^(-1/3 - 1) = (-1/3) *x^(-4/3)

You can not eliminate the 0 from the denominator
The two derivative function are both defined in all R (set of the real numbers) excluding 0, where they are not defined and 0 is a point of discontinuity.

Thanks for the help.

When referring to removing 0 from the denominator, you were talking about the two derivative functions you solved for me, no?

What is the derivative of x^(1/3) and -x(1/3)? I tried finding an equation that eliminated 0 from the denominator, but failed. I also tried using the “cheat” method (x^n ->
nx^(x-1), but couldn’t see how this was implemented.

If possible could you show the algebra steps? I don’t need the calculus notation, but would like to know how it works, not just the answer.

If we start going into historical accuracy, we would have to make disparate units (e.g. German tanks hit at 4, Russian at 3, Japanese at 2).

You are still paying for that J1 against me.

No one has noticed the age on this thread?

32%.

Oh. Well then I guess that could be just changed into drinking extensively in general.

Ten dollars is still a lot for glorified grape juice.

INTJ, thinking at 50%, judging at 56%.

I was more thinking of smoking, gambling, 4 dollar lattes, and 50 dollar bottles of wine.

Anyways, you can’t really teach these kinds of skills to high schoolers because many of them don’t really care about school that much. They don’t pay attention in class, and really don’t learn anything except for the tests, especially the ones who would be at risk of going bankrupt. The smart students, who pay attention, don’t really need this because they are intelligent enough to figure out how to manage their finances on their own, and how to budget.

In essence, the class would be preaching to the choir, and the atheists at the same time. People are either already going to know, or not care.

(I know this from experience.)

Getting rid of expensive, addicting habits would be an easy way to eliminate a large amount of bankruptcy.