IS it possible to get an answer?

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- Mar 23rd 2006, 03:30 AMMaithamDifferentiation and Integration
IS it possible to get an answer?

url=http://www.eyelash.ps/up]http://www.eyelash.ps/up/uploads/448c4941bb.jpg[/url] - Mar 23rd 2006, 05:37 AMCaptainBlackQuote:

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 - Mar 23rd 2006, 05:43 AMCaptainBlack
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 - Mar 27th 2006, 09:51 PMMaitham
Thanks For the answer and i`ll try to do the rest and post the answer to check them for me if they are right, thats if it possible with you.