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I have to solve an ODE with variation of coefficient technique. It's pretty easy but I have no clue what is the first and second derivative of e^ix and e^-ix.

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- #2

arildno

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If i had been a real number, what would the first and second derivatives have been then?

- #3

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e^ix

first

i*e^ix

second

i^2*e^ix

e^-ix

first

-i*e^-ix

second

i^2*e^-ix

p.s. I've read about the Cauchy-Riemann equation but just not sure how to apply it... should I split the exponential in a sin and a cos?

p.s.s. There are probably rules, like exponential function are always derivable or something but I'm not fallowing any complex variables class right now so any insight is appreciated...

first

i*e^ix

second

i^2*e^ix

e^-ix

first

-i*e^-ix

second

i^2*e^-ix

p.s. I've read about the Cauchy-Riemann equation but just not sure how to apply it... should I split the exponential in a sin and a cos?

p.s.s. There are probably rules, like exponential function are always derivable or something but I'm not fallowing any complex variables class right now so any insight is appreciated...

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- #4

arildno

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e^ix

first

i*e^ix

second

i^2*e^ix

e^-ix

first

-i*e^-ix

second

i^2*e^-ix

EXACTLY!

And that is precisely what holds when "i" is a complex/imaginary number as well!

- #5

Gib Z

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[tex]\exp(ix),\,\,\,x\in \mathbb{R},[/tex]

(which is what it looks like you have) then it's what the above two said.

[tex]\exp(iz),\,\,\,z\in \mathbb{Z},[/tex]

you need to be more careful. Let us know if that is indeed what you have.

- #7

mathwonk

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what are you doing in a de course ifm you do not know the derivative of e^z?

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- #9

HallsofIvy

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That is exactly what

That is a result of the very basic fact that the derivative of [itex]e^x[/itex] is [itex]e^x[/itex] (world's easiest derivative!) and the chain rule.

- #10

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[tex]\frac{d}{dx}(e^{jx})=je^{jx}[/tex]

[tex]\frac{d^2}{dx^2}(e^{jx})=-e^{jx}[/tex]

- #11

mathwonk

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