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well.. depends on what substitution it is i think 2u ppl shud be able to differentiate y=e^(x lnx) or maybe simpler ones like y=e^sinx but the trick with this q is to get y=x^x y=e^(x lnx) and that shudnt be in 2u

doesnt work against me hihihihihihihi

HK attack! if u dont want to use the method they do in Hong Kong, y=x^x y=e^(x lnx) if u=x lnx, du/dx = 1 lnx + x 1/x = lnx + 1 y=e^(x lnx) y = (lnx + 1) e^(x lnx) hihihihihi *flee!* *comes back* anyway u wont get this kind of question *steal!* *flee!*

lol this is funny

quadrilateral is anything with four sides or four points

u also solve dy/dx = f(x).. like dy/dx = sin(x) and so on

he went to HK so its unlikely that he'll be answering maths q from over there unless he loves maths more than u do ;)

he/she's not FinalFantasy lol

if (pi^4 + pi^5)^1/6 was e then people wouldn't have invented a new symbol "e" so its just an approximation

u dont sub t=1 let Φ = f (tx,ty) = t^n f(x,y) dΦ/dt = (∂f/∂tx)*(dtx/dt) + (∂f/∂ty)*(dty/dt) = n*t^(n-1) f(x,y) (∂f/∂tx)*x + (∂f/∂ty)*y = n*t^(n-1) f(x,y) (∂f/∂tx)*tx + (∂f/∂ty)*ty = n*t^n f(x,y) (∂f/∂tx)*tx + (∂f/∂ty)*ty = n*f(tx,ty) (∂f/∂u)*u + (∂f/∂v)*v = n*f(u,v) (∂f/∂x)*x +...

lol I just found out that the extension 2 files are not on the main server. I uploaded them just then but it's still not working u can use this one for now: http://www.geocities.com/nsweducationalservice/maths/

I have a list here: http://forum.aatraders.net/alp_view.php?t=anime&u=somebody If you're interested, give me a holla as well :D I'll keep updating this list. funniboi, ur fumoffu looks right If anyone wants to know about the number of episodes and the fansub groups, this site is...

assuming y=In(x) refers to y=ln(x), u wont see this in 2U exam either ln(x) is not defined at x=0 if y=In(x) refers to y=integral of x dx then it can be in 2U

if y=ln(x) and its rotated about x-axis: Install Maple Open it Type: VOLUME:=evalf(pi*value((simpson((log(x))^2, x=1*10^(-99)..3, 10000)))) ; and it'll find the volume using Simpson rule, which is what u asked. (edited to make the initial x value smaller and to fix up the command)

60 km/h = 60000 m/h = 60000/3600 m/s = 50/3 m/s Do the usual integration stuff and you'll find: x = 50/3 cos@ t y = -4.9t^2 + 50/3 sin@ t + 0.9 put (x=18, y=3) into the above equations: 18 = 50/3 cos@ t t = 1.08 / cos@ 3 = -4.9t^2 + 50/3 sin@ t + 0.9 3 = -4.9(1.08 / cos@)^2 + 50/3...

thats too big if it says The page cannot be displayed.

I used Rikku's Mix ability to combine 2 rare healing items.

lol definitely not... its not even used in any hsc textbooks this one is more like it: INT cos^3 x dx = INT cos x * cos^2 x dx = INT cos x * (1-sin^2 x) dx = 1 - 1/3 sin^3 x + C hihihi EDIT: its sin x - 1/3 sin^3 x + C

the expansion of cos3@ is rarely used in 3U as well... pi times (integral of cos 2x dx from x=pi/6 to x=pi/4) should give you pi/4(2-root(3)) That's right =p

1. = cos x * cos^2 x = cos x * (1-sin^2 x) the derivative of sin x is cos x let u = sin x INT cos x * (1-sin^2 x) * dx = INT du * (1-u^2) 2. Vol = pi*INT (cos x)^2 dx - pi*INT (sin x)^2 dx = pi*INT (cos^2 x - sin^2 x) dx = pi*INT (1 - 2sin^2 x) dx = pi*INT (1 - 2[1/2 - 1/2...