Maths genius?
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12:10 Sun 23 Dec 07 (GMT) [Link]
There is no need to make the excercise more complicated, you need to solve it !!
ruddan_06_ said:
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)
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)
There is no need to make the excercise more complicated, you need to solve it !!
12:30 Sun 23 Dec 07 (GMT) [Link]
lim (e^2x - 1)^(1/lnx)
x-> 0
_>
Uh-huh, but (e^2x - 1)^(1/lnx) isn't of the form f(x)/g(x)...
x-> 0
_>
Uh-huh, but (e^2x - 1)^(1/lnx) isn't of the form f(x)/g(x)...
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13:15 Sun 23 Dec 07 (GMT) [Link]
Sed it wrong
Edited at 19:16 Sun 23/12/07 (GMT)
Edited at 19:16 Sun 23/12/07 (GMT)
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13:17 Sun 23 Dec 07 (GMT) [Link]
There is no need to make the excercise more complicated, you need to solve it !!
Jeesus Man ! !
SOlve it !?!?!?!
No thanks.....id rather jump in a house full of bee's for a year !
Edited at 19:17 Sun 23/12/07 (GMT)
osullivanr said:
ruddan_06_ said:
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)
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)
There is no need to make the excercise more complicated, you need to solve it !!
Jeesus Man ! !
SOlve it !?!?!?!
No thanks.....id rather jump in a house full of bee's for a year !
Edited at 19:17 Sun 23/12/07 (GMT)
15:04 Sun 23 Dec 07 (GMT) [Link]
No?
I'll post the answer tomorrow
Edited at 21:09 Sun 23/12/07 (GMT)
clooneman said:
lim (e^2x - 1)^(1/lnx)
x-> 0
_>
Uh-huh, but (e^2x - 1)^(1/lnx) isn't of the form f(x)/g(x)...
x-> 0
_>
Uh-huh, but (e^2x - 1)^(1/lnx) isn't of the form f(x)/g(x)...
No?
I'll post the answer tomorrow
Edited at 21:09 Sun 23/12/07 (GMT)
14:50 Wed 9 Jan 08 (GMT) [Link]
Haha oops
clooneman said:
Then... shouldn't it have begun with 0?
Anyhow, my sum is:
142857 x 7
Anyhow, my sum is:
142857 x 7
Haha oops
17:38 Wed 9 Jan 08 (GMT) [Link]
I'm meant to come up with a result for that derivative for ye
16:35 Mon 14 Jan 08 (GMT) [Link]
soz lads i was off to RuneScape, i'm going to bed now, i'll answer tomorro
15:46 Sat 23 Feb 08 (GMT) [Link]
I dragged it off to another forum for their perusal and the various answers I got were:
(sin 3x)^(2-x) {3(2-x)(cot 3x) - Ln(sin 3x)}
and
(sin 3x)^(2-x) * (- ln sin 3x + (2-x)* 3cos3x / sin 3x)
and
(3(2-x)Cot(3x)-Log(Sin(3x)))(Sin(3x)^(2-x))
All four of these answers are the same.
So, no-one got it... Another one for us, osullivanr?
osullivanr said:
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
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
I dragged it off to another forum for their perusal and the various answers I got were:
(sin 3x)^(2-x) {3(2-x)(cot 3x) - Ln(sin 3x)}
and
(sin 3x)^(2-x) * (- ln sin 3x + (2-x)* 3cos3x / sin 3x)
and
(3(2-x)Cot(3x)-Log(Sin(3x)))(Sin(3x)^(2-x))
All four of these answers are the same.
So, no-one got it... Another one for us, osullivanr?
17:05 Mon 21 Apr 08 (BST) [Link]
*walks into the forum*
*says WTF somebody tried to solve it*
*looks at his answer*
*cries cos it ain't rite *
*decides to give a hint*
*says the answer is e*
*hopes ruddan will be able to solve it now
*
Gl fella
*says WTF somebody tried to solve it*
*looks at his answer*
*cries cos it ain't rite *
*decides to give a hint*
*says the answer is e*
*hopes ruddan will be able to solve it now
*
Gl fella
17:08 Mon 21 Apr 08 (BST) [Link]
*returns to RuneScape*
*W8s for his first fp birthday*
*W8s for his first fp birthday*
18:23 Sun 4 May 08 (BST) [Link]
The answer isn't e!
Emmm, he does know it; like I said, his answer is the same as those I got from a different forum on a different website!
osullivanr said:
*decides to give a hint*
*says the answer is e*
*says the answer is e*
The answer isn't e!
ruddan_06_ said:
He doesnt even no it
Emmm, he does know it; like I said, his answer is the same as those I got from a different forum on a different website!
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Maths genius?
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