I'll startwith and easy one:

(492+ 652)/4

Once you have answered it then you have to put a sum up for the next person to answer

A post an answer

a post a question

I'm gonna do this in my head...

1144/4
is...
286!!!

and your sum cloone is???

ur go now

Another good maths thread would be a guess the sequence one. ie. i put :

{1, 4, 12, 40, 84, 172...}

And the person has to guess the sequence

= 2n+4 in this case

Then... shouldn't it have begun with 0?

Anyhow, my sum is:

142857 x 7

999999?

right, now we'll take it to a higher level:

dy/dx (sin 3x)^(2-x)

ift no one can solve it before saturday, i shall post an easier one

Oooh, nasty... What was that chain rule again?

y = (sin 3x)^(2-x)

So...
dy/dx (sin 3x)^(2-x)
=
d{(sin 3x)^(2-x)}/d{(sin 3x)}
times
d{(sin 3x)}/d{3x}
times
d(3x)/dx
=
(2-x)(sin 3x)^(1-x).(cos 3x).3
= (6-3x)(cos 3x)(sin 3x)^(1-x)

Is that right?

dy/dx = (dy/dv)(dv/dx), innit? (

Perhaps I'd better make that look a bit more accessible

Let u = 3x
Let v = sin u
y = v^(2-x)

So...

u = 3x
du/dx = 3

v = sin 3x = sin u
dv/dx
= dv/du x du/dx
= 3(cos u)

y = (sin 3x)^(2-x) = (sin u)^(2-x) = v^(2-x)
dy/dx
= dy/dv x dv/dx
=...

And that's where I get stuck on a technicality, cos:
y = v^(2-x)
so dy/dv is hampered by the x, isn't it? Anyone?

Are we doing algebra?

I havent been tought that that yet!!

y = n^x
dy/dx =... (log x) n^x or something

Edited at 17:19 Thu 20/12/07 (GMT)

No, here it is:

y = n^x
dy/dx = (log n) n^x
And, if y = n^v where v is some function of x,
then dy/dx = (log n)n^v · dv/dx.

So, starting again:
y = (sin 3x)^(2-x)

Let u = 3x
Let v = sin u
Let w = 2-x
Let y = v^w

So...

u = 3x
du/dx = 3

v = sin 3x = sin u
dv/dx
= dv/du · du/dx
= 3(cos u)
= 3(cos 3x)

w = 2-x
dw/dx = -1

y = (sin 3x)^(2-x) = (sin u)^(2-x) = v^(2-x) = v^w
dy/dx
= log v · v^w · dw/dx · dv/dx (I hope)
= (log sin 3x)((sin 3x)^(2-x))(-1)(3)(cos 3x)
=-3(log sin 3x)((sin 3x)^(2-x))(cos 3x)

That can't be right, can it?

Hmmm... I know... you're gonna say the way to do it is

(sin 3x)^(2-x)
= (sin 3x)²/(sin 3x)^x
and use the quotient rule, aren't ya!!!

clooneman- was i right?

Personally I lack the ability to solve such problems, however, I do have the ability to find out

I asked on another site and a lot of people came up with a lot of answers...

I chose this one...

You was!!!

Really? And no log in there? Hmmmmm.....

Errr...Arrr.......Ee......Eekk....

*Slowly Puts hands in pockets.......and veryyyy slowly turns around and walks away...hoping not to ever to bump into such a horrible thread again !*

I personally dont think it is possible...

Nowhere I have searched has gone, yeah here is the definitive answer.

It would be helpful if osullivanr replied