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

1. Feb 8, 2006

### 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?

2. Feb 8, 2006

### quasar987

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)

Last edited: Feb 8, 2006
3. Feb 8, 2006

### 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?

4. Feb 8, 2006

### UrbanXrisis

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

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

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

5. Feb 9, 2006

### VietDao29

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:
$$ia e ^ {iaz} = ia (\cos (az) + i \sin (az)) = ia \cos (az) - a \sin (az)$$