bit of maths help

:confused:

If you're going to post a thread asking for help, why not try actually mentioning what it is you want help with. This is the second thead in two days in which you've asked for help with your homework, without actually explaining wtf it is you want.

You've just posted a few formulas here (which are totally inconsistent by the way).

Also, I suggest you work out how to use parentheses to correctly express your formulas.
 
Looks like a fine l'Hopital's rule question. We have an indeterminate form 0/0, so we differentiate:

(3e^(3x) - 3) / 2x

Same deal, we have a 0/0 form, so we differentiate again:

9e^(3x) / 2

Taking the limit, we get the limit 9/2. Alternatively, we can interpret e^(3x) by its Maclaurin series, which is:

e^(3x) = 1 + 3x / 1! + 9x^2 / 2! + 27x^3 / 3! + ...

So the expression in the limit is:

[(1 + 3x + 9x^2 / 2 + 27x^3 / 6 + ...) - 1 - 3x] / x^2
= [9x^2 / 2 + 27x^3 / 6 + ...] / x^2
= 9 / 2 + 27x / 6 + ...

Taking the limit as x approaches 0, we get 9 / 2 as before.
 
f(x)=e^x3-1 . (3x+5)^3
f ' (x)= 3e^3x-1 . 9x(3x+5)^2

i thought when you F' an exponent
e^3x-1 = e^3x-1 . 3 = 3e^3x-1

next one

e^3x^2 -1= e^3x (3X^2)= 3x^2 e^3x-1
 
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Why are you using the equals sign? Youre just doing chain rule differentiation?

f(x) = e^(3x^2)

f'(x) = 6xe^(3x^2)
 
your equations still don't make sense, try writing in maths that actually makes sense!!

if f(x) = e^(3*x^2)
then f'(x) = 6x*e^(3*x^2)

when differentiating a function of a function, f(g(x)), the general formula is g'(x)*f'(g(x))
 
17 posts and it's still completely unclear what you're even trying to do.

Try reading what you write back to yourself. I think you have a problem with coherency. How can anyone help if your formulas don't make sense, and if you don't explain what you're trying to do. Also, use parenthesis to remove the ambiguities of writing formulas in text. Or better yet, use equation editor and take a screengrab of the result.
 
mattheman said:
e^3x-1 = e^3x-1 . 3 = 3e^3x-1

(3x+5)^3 = 9x(3x+5)^2

so
f(x)=e^x3-1 . (3x+5)^3
f ' (x)= 3e^3x-1 . 9x(3x+5)^2
Click to expand...

Also this would be

f(x) = e^(3x-1).(3x+5)^3

f'(x) = 3e^(3x-1).(3x+5)^3 + e^(3x-1).3.3.(3x+5)^2

edit:

http://forums.overclockers.co.uk/showpost.php?p=14346858&postcount=22

Did you do A-level Maths? This is like the first thing you learn.

mattheman said:
(masters) <--- hope so
degreeCertificateDiploma
A-Level’s
8 GCSE’s
IAQ
CMI, Chartered Management Institute.
MCSA 2003 .
CCNA,
C&G 2 of them.
SFA
(CFA) <------hope so.
imc <-------going to so soon.
Click to expand...

Also in this thread http://forums.overclockers.co.uk/showthread.php?t=18020834&highlight=username_mattheman

you say that you got a job as an analyst in fixed income and derivative markets and you can't even differentiate?
 
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marc_howarth said:
your equations still don't make sense, try writing in maths that actually makes sense!!

if f(x) = e^(3*x^2)
then f'(x) = 6x*e^(3*x^2)

when differentiating a function of a function, f(g(x)), the general formula is g'(x)*f'(g(x))
Click to expand...

how did you get f'(x)=6x*e^(3*x^2)

i thought Y=e^(3x^2) would be y' e^(3x^2) . 3 = y' 3e^(3x^2)
 
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