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 [quote="CrystyB"]I'll have to make a note of this in me "Dearest and Beloved Problems"! [img]http://www.greylabyrinth.com/Forums/smile.gif[/img] My answer would have been (at any hour of the day, after no-matter-how-many math experiences for the day): [cos(alpha(x))]'=[-sin(alpha(x))]*[alpha'(x)] and [sin(alpha(x))]'=[cos(alpha(x))]*[alpha'(x)] so f'(x)=i*sin(-ix)+i*(-i)*cos(-ix)=cos(ix)-i*sin(ix). And just now my experiences for the day would make a difference: i wonder how soon would i have noticed that f'=f so f=ct*e^x... edited on 11-18-2000 and [This message has been edited by CrystyB (edited 11-30-2000).][/quote]
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CrystyB
Posted: Sat Nov 18, 2000 5:22 pm    Post subject: 1

I'll have to make a note of this in me "Dearest and Beloved Problems"!

My answer would have been (at any hour of the day, after no-matter-how-many math experiences for the day):

[cos(alpha(x))]'=[-sin(alpha(x))]*[alpha'(x)]
and
[sin(alpha(x))]'=[cos(alpha(x))]*[alpha'(x)]

so f'(x)=i*sin(-ix)+i*(-i)*cos(-ix)=cos(ix)-i*sin(ix).

And just now my experiences for the day would make a difference: i wonder how soon would i have noticed that f'=f so f=ct*e^x...

edited on 11-18-2000 and

[This message has been edited by CrystyB (edited 11-30-2000).]
Ghost Post
Posted: Thu Oct 19, 2000 1:18 pm    Post subject: 0

Please correct me if I'm wrong...

e^iz = cos(z) + i*sin(z)
let z = -ix
so...
f(x) = e^(i*(-ix))
i^2 = -1
f(x) = e^x therefore
f'(x) = e^x

The derivitate is the same as the function.

[This message has been edited by Matt (edited 10-19-2000).]
Alfie
Posted: Thu Oct 19, 2000 4:44 am    Post subject: -1

I am putting this here because it belongeth not to any other place.

f(x)=cos(-xi)+sin(-xi)i
Find the first derivitive of f(x). What does this mean and why would I find it interesting?