$\displaystyle f(t) = \frac{pqe^t}{(1-qe^t)^2}$

$\displaystyle \frac{d}{dt}f(t) = \frac{(pqte^t)(1-qe^t)^2 -2(1-qe^t)(-tqe^t)(pqe^t)}{(1-qe^t)^4}$ $\displaystyle = \frac{[(pqte^t)(1-qe^t)][(1-qe^t)+2(pqe^t)]}{(1-qe^t)^4}$

is this right?

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- Mar 23rd 2008, 11:11 PMlllllderivative question
$\displaystyle f(t) = \frac{pqe^t}{(1-qe^t)^2}$

$\displaystyle \frac{d}{dt}f(t) = \frac{(pqte^t)(1-qe^t)^2 -2(1-qe^t)(-tqe^t)(pqe^t)}{(1-qe^t)^4}$ $\displaystyle = \frac{[(pqte^t)(1-qe^t)][(1-qe^t)+2(pqe^t)]}{(1-qe^t)^4}$

is this right? - Mar 23rd 2008, 11:24 PMTheEmptySet
No

in your first derivative

$\displaystyle pqe^t$ the derivative is $\displaystyle pqe^t$

remember $\displaystyle \frac{d}{dx}e^t=e^t$ not $\displaystyle te^t$

Good luck. - Mar 23rd 2008, 11:31 PMlllll
Thanks, I don't know how I forget about deriving e.