Q:

$\frac{\partial}{\partial t} e^{\lambda e^{it-1}}$ and $\left( 1-\frac{it}{\alpha}\right)^{-\beta}$

$\frac{\partial}{\partial t} e^{\lambda e^{it-1}}$
$\frac{\partial}{\partial t}\left( 1-\frac{it}{\alpha}\right)^{-\beta}$
I calculated the characteristic functions of the Poisson and Gamma distributions to be the above function and am trying to calculate expectation and variance from them.
It has been years since I took calculus and I am pretty unsure how to take the derivatives. 
I do know that I can find $E(X)$ by taking the first derivative of the characteristic function and evaluating it at t=0, (and similar for higher moments) and I know what the answers for expectation and variance are for these distributions, but I am struggling with the calculus.
Any help is appreciated

A:

You can break this into a chain rule by writing it as the composition of two functions. Let $f(x)=e^x$ and $g(t)=\lambda e^{it-1}$. Then $f(g(t))=e^{\lambda e^{it-1}}$. 
The chain rule says $\frac{d}{dt}(f(g(t)))=f'(g(t))\cdot g'(t)$. The derivative $f'(x)=e^x$ and you need another chain rule for $g'(t)=\lambda ie^{it-1}$. Thus $\frac{d}{dt}\left(e^{\lambda e^{it-1}}\right)=\lambda ie^{it-1}e^{\lambda e^{it-1}}$
Does this help? You should just use the chain rule for the second one as well and the fact that $\frac{d}{dt}(f(t))^\beta=\beta f(t)^{\beta-1}\cdot f'(t)$

