bit of maths help

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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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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ghost101 said:
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.



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?
Click to expand...

iam trying to remember this stuff.
 
marc_howarth said:
here are some other examples that might help you understand

f(x) = sin(3*x), f'(x) = 3*cos(3*x)

f(x) = e^(sin(x)), f'(x) = cos(x)*e^(sin(x))

basically, you have a function inside another function,

e being the exponential function, 3x^2 being the function side the exponential function

is the case of f(x) = e^(3*x^2), this can be written as the 2 seperate function such that f(x) = g(h(x)). g(x) = e^x and h(x) = 3*x^2

using the general formula f'(x) = h'(x)*g'(h(x))
g'(x) = e^x
h'(x) = 6*x

another way of putting it would be to differentiate the bit inside, and this is what you multiply the initial function by once you have differentiated that
Click to expand...

so y=e^3x = y'=3e^3x so y'' 9e^3x
this answer is right just checked it .
 
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