Maths (core 3)

by Ollz • 9 Nov 2008, 17:36 • Journals
i'm trying to find the gradient of the graph y = 2e^(3x+1) at the point (0,2e)

what the fuck?

thankyou.

EDIT: solved
Comments
23
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ED 9 Nov 2008, 17:37
read your book np.
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Ollz 9 Nov 2008, 17:38
i had time off, so i have no notes on this stuff :(
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ED 9 Nov 2008, 17:40
ask your meights.
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defOK 9 Nov 2008, 17:39
you have to search harder :<
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eiM 9 Nov 2008, 17:42
f'(x) = 6*x^(3x+1) * ln(x) + 2*(3x+1)*x^(3x)
f'(0,2e) = 1,211712728

If I didnt mistranslated or mistype in my CAS :D
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Ollz 9 Nov 2008, 17:50
i have trouble following that, but thanks if it's right :D
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eiM 9 Nov 2008, 17:53
well there are rules to make f'(x) ... you gotta know them and use them :p Sry but i dont know the words in english^^
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spiROZE 9 Nov 2008, 18:08
f(x)=e^(ln(f(x))

you have to make it with that rule :)
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MindGamee 9 Nov 2008, 17:44
gl w dat shit
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im a tree 9 Nov 2008, 17:45
which chapter/page is it in the c3 book and i might take a look for you
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Ollz 9 Nov 2008, 17:48
only notes :<

i thought you'd know it off by heart anyway :D
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im a tree 9 Nov 2008, 17:48
i havent even done it yet, c3 isnt until like January
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spiROZE 9 Nov 2008, 17:47
is izi, you have to play with ln and e to make it :)
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panics 9 Nov 2008, 18:27
MATHS IS GAY OLLAOLWOALWOLALOL
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Mayni 9 Nov 2008, 18:43
Im so glad ive finished my alevel maths now, just thinking of this stuff gives me a headache :DD


e: Think you need to take logs of both sides, and take it from there. Youve got values for x/y as well, so use them if you can.

Tbh that might be quite wrong, i really cant remember c3/c4 anymore :<
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_gamboo 9 Nov 2008, 18:46
fu
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Mayni 9 Nov 2008, 18:50
<3, still on neptulon btw?
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_gamboo 9 Nov 2008, 18:54
yeah, not playing atm though, guild went inactive till the expansion and since then I dont feel like playing :D
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Mayni 9 Nov 2008, 18:54
Going to play wotlk?
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_gamboo 9 Nov 2008, 19:06
doubt it, since friends stopped playing the game aint fun :(
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_gamboo 9 Nov 2008, 18:47
what part do you get stuck on?
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_gamboo 9 Nov 2008, 19:24
Differentiating gives the gradient function.

y=e^kx -> dy/dx=ke^kx
so
y=2e^(3x+1) -> dy/dx= 6e^(3x+1)

Then for finding the gradient at a point you simply place the x,y value(s) needed from the co-ordinate of the point into the function.

6e^(3(0)+1) = 6e =16.309....
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Ollz 9 Nov 2008, 19:28
thankyou, i'm sorted now =]

didn't know what to do because i'd never had more than one term in the power :p
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