Differentiate e^(a^2 z) with respect to z

[UrbanXrisis]

I am to differentiate with respect to z, where a is independent of z... (am assuming that that a is a constant?)

a. e^{a^2 z}=diff=>a^2 e^{a^2 z}

b.e^{ia z}=cos(az)+isin(az)=diff=>-sin(az)+icos(az)

c. (e^{-i z})^2=cos(2z)-isin(2z)=diff=>-sin(2z)-icos(2z)

d. e^{-iz^2}=diff=>-sin(z^2)-icos(z^2)

e. e^{az}+e^{-az}=diff=>ae^{az}+-ae^{-az}

do these look alright?

 

[quasar987]

(am assuming that that a is a constant?

Whatever it is, it does not depend on z, so when z changes, a does not change. That's the definition of "constant".

Double check b,c,d; you're forgetting the derivative of the interior (chain rule).

note: you don't need to write e^ix in temrs of cis(x) b4 taking the derivativ; complex exponentials differentiate like regular ones, i.e.

(e^iax)' = ia(e^iax)

 

[UrbanXrisis]

for b.
-azsin(az)+azicos(az)

for c.
2isin(2z)-2cos(2z)

for d.
-4z(icos(z^2)+sin(z^2))+2(cos(z^2)-isin(z^2))

how does it look?

 

[UrbanXrisis]

let me correct myself:
for b:
-asin(az)+aicos(az)

for c:
-2sin(2z)-i2cos(2z)

for d:
-2z(sin(z^2)+icos(z^2))

 

[VietDao29]

let me correct myself:
for b:
-asin(az)+aicos(az)

for c:
-2sin(2z)-i2cos(2z)

for d:
-2z(sin(z^2)+icos(z^2))

Looks good. But as quasar987 said before, just try to differentiate it without changing to cis function.
(e ^ {iaz})' = ia e ^ {iaz}, which is a lot easier, right? This answer is exactly the same as yours:
ia e ^ {iaz} = ia (\cos (az) + i \sin (az)) = ia \cos (az) - a \sin (az)