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Test Ourselves: P3 watch

1. (Original post by Ralfskini)
No, keep trying.
ok..
is it

(x^x^x)(x^(x-1) + (x^x)ln x + (x^x)(ln x)^2) ?
2. (Original post by Ralfskini)
No, keep trying.
I get this nasty contraption:

( x^x^x ) ln(x^x^x) [ 1 + lnx + 1/(xlnx) ]
3. what about this question from last years summer paper, its easy, but i have cant remember how to do it

find the cartesian equation and sketch the curve from the following parametric equations

x = 3cost y=cos2t
4. (Original post by kimoni)
ok..
is it

(x^x^x)(x^(x-1) + (x^x)ln x + (x^x)(ln x)^2) ?

Yes, that's what I got (although I did make up the question)
5. (Original post by scottyoneill)
what about this question from last years summer paper, its easy, but i have cant remember how to do it

find the cartesian equation and sketch the curve from the following parametric equations

x = 3cost y=cos2t
y = cos2t = 2cos^2(t)-1 = 2(x/3)^2 - 1
y = (2/9)x^2-1

which is a parabola going though (0,-1) and (3/sqrt(2),0), (-3/sqrt(2),0).
6. (Original post by kimoni)
ok..
is it

(x^x^x)(x^(x-1) + (x^x)ln x + (x^x)(ln x)^2) ?
could you show me step by step?
i've no idea how i can solve it..
7. here's one for all u then:diffentiate y=x^x^x^x^x^x^x^x^x^x^x.......(g oes on forever)
8. (Original post by ralphk)
could you show me step by step?
i've no idea how i can solve it..
y=x^x^x then lny=ln(x^x^x)=(x^x)lnx
9. (Original post by Mrm.)
Just use integration by parts with u=x and v´=Cos 2x
ie u´=1 and v = 1/2 Sin 2x

therefore

I xCos2x dx = x . 1/2 Sin 2x - I I/2 Sin 2x dx

= 1/2 x 2 Sin x . Cos x + 1/4 Cos 2x + c

= 1/2 x 2 Sin x.Cos x + 1/4 (1 - 2 Sin^2 x) +c

= 1/2 x 2 Sin x.Cos x + 1/4 - 1/2 Sin^2 x +c

= 1/2 x 2 Sin x.Cos x - 1/2 Sin^2 x + k

= 1/2(sinx(2xcosx-sinx) + k
where did the highlighted cosx come from?
10. identity: sin 2x = 2 sinx.cosx
11. (Original post by Mrm.)
identity: sin 2x = 2 sinx.cosx
oh yea i see... thanks!

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