IS it possible to get an answer?
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Lets do the first derivative problem:Originally Posted by Maitham
Differentiate:
$\displaystyle
f(t)=\exp(\mu t + \sigma^2t^2/2)
$
Here we need the chain rule:
$\displaystyle
\frac{d}{dt}g(h(t))=\frac{dh}{dt}(t) \times \frac{dg}{dt}(h(t))
$,
so put $\displaystyle h(t)=\mu t + \sigma^2t^2/2$ and $\displaystyle g(t)=exp(t)$.
Then:
$\displaystyle
\frac{d}{dt}f(t)=[\mu+\sigma^2 t][\exp(\mu t + \sigma^2t^2/2)]
$
RonL
Now lets do the first of the integrals:
$\displaystyle
\int_0^{\infty} \frac{1}{\beta}e^{-x/\beta}dx
$
The integrand is the derivative of $\displaystyle -e^{-x/\beta}$ so we know
(probably from the fundamentaly theorem of calculus) that:
$\displaystyle
\int_0^{\infty} \frac{1}{\beta}e^{-x/\beta}dx=\left[-e^{-x/\beta}\right]_0^{\infty}=-0+1=1
$
RonL