Differentiating an exponential with a complex exponent

[Zacarias Nason]

Hello, folks. I'm trying to figure out how to take the partial derivative of something with a complex exponential, like
\frac{\partial}{\partial x} e^{i(\alpha x + \beta t)}
But I'm not really sure how to do so. I get that since I'm taking the partial w.r.t. x, I can treat t as a constant term and thus pretend it's something like
\frac{\partial}{\partial x} e^{i(\alpha x +\beta)}
But then my confusion comes from me not being able to separate the exponent into some suitable form like
e^{\alpha + \beta i}
I guess I could separate it into two separate ones, like
e^{i\alpha x}e^{i\beta}

How should I deal with this, any pushes in the right direction?

 

[fresh_42]

You should apply the chain rule of differentiation to $e^{f(x)}$.

 

[Zacarias Nason]

Thanks, let me take a shot.

\frac{\partial}{\partial x}e^{i(\alpha x +\beta t)} = e^{i (\alpha x + \beta t)} \cdot \frac{\partial}{\partial x}[i(\alpha x + \beta t)]= i \alpha e^{i (\alpha x + \beta t)}

 

[Zacarias Nason]

Is this correct, I guess? I got the same thing by the product rule.

Edit: it must be, because when I applied this bit to the larger problem I was working on I got it right! Thanks so much!

 

[fresh_42]

Is this correct, I guess? I got the same thing by the product rule.

Yes, except you lost the $t$ in your product rule. It's going to zero (in one term) as it is viewed as a constant by partial differentiation but you must not just drop it before and then again pull it out of the hat again.

 

[Zacarias Nason]

Uh oh, I don't get what you just said, let me work it out again both ways and see where I'm making the mistake.

 

[fresh_42]

\frac{\partial}{\partial x} e^{i(\alpha x + \beta t)}
But I'm not really sure how to do so. I get that since I'm taking the partial w.r.t. x, I can treat t as a constant term and thus pretend it's something like
\frac{\partial}{\partial x} e^{i(\alpha x +\beta)}

$t$ has gone.
And you said you got the same result by product and by chain rule, so there is a $t$ again of the exponent in the result.

 

[Zacarias Nason]

Oh, ok.