Omg i feel physically sick inside and my head goes all dizzy and i newarly puke up and my eyes bulge out and i get a high fever and my hairs stand on end.....Whenever i scroll up and see all those.......number things

well osullivanr has got til tomorrow as he said he will be on then

Isn't it simple differentiation?


so dy/dx = (2-x)(sin*3)x^(2-x)-1

Or sumthin like that.

Edited at 19:19 Fri 21/12/07 (GMT)

No, ive worked it out

dy=(2-3*)x^(3-x)-1^
dx = 2^(3-*)x*(3-^2*)*+3^

There are the formulas....u shud no the rest

*Walks into forum*

*Immediately turns round and walks back out*

give me til after the marathon and ill do it

Bah I cant find my log tables...when I do Ill do it

Lmao

*stulovescat repeats this action*

ah here we go:

dy/dx (sin 3x)^(2-x)
= dy/dx e^(ln(sin3x)^(2-x))
= e^(ln(sin3x))^(2-x) . dy/dx (ln(sin3x)^(2-x))
= (sin 3x)^(2-x) . (dy/dx((2-x) . ln (sin3x))
= (sin 3x)^(2-x) .(-1(ln (sin3x)) + (2-x) (dy/dx ln (sin3x))
= (sin 3x)^(2-x) .(-ln (sin3x) + (2-x) ((dy/dx 3x .cos3x)/sin3x)
= (sin 3x)^(2-x) (3(2-x) cot3x -ln(sin3x))

Nice to see so many people broke their heads

You people are lucky not to come in touch with my teacher of maths :O

How da -censored- can u use log here?????




Edit: my 100th post

Edited at 01:41 Sun 23/12/07 (GMT)

Ok the next one:

lim (e^2x - 1)^(1/lnx)
x-> 0
_>

Edited at 01:44 Sun 23/12/07 (GMT)

Well, you use limits, which people often confuse with logs...easy mistake! If I had found my log tables I would have done it...I didnt know have the top off my head what sin3x differentiated to!

Edited at 01:49 Sun 23/12/07 (GMT)

sin 3x differentiates to 3 cos 3x, I do believe...?

Let's see, this limit is a bit gnarly, but can I take a guess and say it's undefined?

EDIT: Right, I dragged it into Excel and examined the value of the expression with ever decreasing and the answer is approximately 2.66. However I looked at the graph and it seems that it tails off towards the exact value before becoming undefined. Doesn't this mean that we can tie the derivative of in somehow? Cos as x approaches 0, the derivative does the same? Hmmmm....

Edited at 05:45 Sun 23/12/07 (GMT)

I'll post the answer next Wednesday;

With the limit, u got to use L'Hopital:

lim (f(x)/g(x)) = 0/0 or infinity/infinity
x->a
=
lim (f'(x)/g'(x))
x->a

I use ln, i don't use limits

--> loge (x) = ln (x)

^_^

Lmao

*Stella follows suit as mass walk out begins*

....Yikes

zzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

*wakes up and thinks "AHAAAAAAAAAAAAA* got it"

Nope just a bad dream

zzzzzzzzzzzzzzzzzzzzzzzzzzzz

sin/coz = 3x(2x^x3)-(cosx3x^)
3x(2x^x3)-(sinx3x^)x(g/^3x^3dis)g/x2^log
(3xg/^x3^/g)
So if ther the log/cos and sin we have to find out the rest which shud a little like this.

(3xg/3^)-(3x4/g^+d/g3^cos) > (4xf/1^)-+(4x3/x^+g/g5^log)
Which al shud come to this
= x2(3^x/g/x^2)x(sinxcos)x(3^cg+2/g^3x)(4g/x^/g)(cosxsin)=(4x^x/g/x3^+r-4x3)