# Differentiating an exponential with a complex exponent

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

• Zacarias Nason
Thanks, lemme take a shot.

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

• fresh_42
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!

Last edited:
fresh_42
Mentor
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.

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
Mentor
$$\frac{\partial}{\partial x} e^{i(\alpha x + \beta t)}$$
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.