What are the steps to take the derivative of this function? This calculus derivation is giving me an extremely difficult time. I am having trouble understanding how to manipulate the $e^t$. The function is
$$P*{e}^t/[(1-q)*{e}^t]$$
I want to take the derivative with respect to $t$ so $d/dt$. The answer to this problem is 
$$[pe^t(1-qe^t)+qe^tpe^t]/(1-qe^t)$$
Any help with basic understanding of how to differentiate these functions would be greatly appreciated. Thank you in advance.
 A: Based on the answer given, I believe the problem was misstated; most likely it is asking to find the derivative of the function
$$f(t) = \frac{pe^t}{1-qe^t}$$
To calculate the derivative, we use the quotient rule:
$$ \frac{df}{dt} = \frac{(1-qe^t)\frac{d}{dt}pe^t - pe^t\frac{d}{dt}(1-qe^t)}{(1-qe^t)^2}$$
Our task is then to determine what $\frac{d}{dt} e^t$ is. Any calculus textbook will state that it is equal to itself, but proving this requires a more sophisticated definition of the exponential function: as the inverse of the natural logarithm, which is defined as $ln(x) = \int_1^x \frac{dt}{t}$.
From the Fundamental Theorem of Calculus, $\frac{d}{dt} ln(t) = \frac{1}{t}$. Now consider that $ln(e^t) = t$ (since the functions are inverses). Taking the derivatives of both sides with respect to $t$,
$$\frac{d}{dt} ln(e^t) = \frac{1}{e^t} \frac{d}{dt} e^t = 1$$
which implies that $\frac{d}{dt} e^t = e^t$.
Returning to our original problem, we obtain
$$\frac{df}{dt} = \frac{(1-qe^t)pe^t - pe^t(-qe^t)}{(1-qe^t)^2} = \frac{pe^t-pqe^{2t} + pqe^{2t}}{(1-qe^t)^2} = \frac{pe^t}{(1-qe^t)^2}$$
where I used the property that $e^te^t = e^{2t}$.
A: Please observe that 
$P*{e}^t/[(1-q)*{e}^t] = pe^t / [(1 -q)e^t] = p / (1- q), \tag{1}$
whence
$(pe^t / (1 - q)e^t])' = 0, \tag{2}$
assuming $P = p$ and $p, q$ are constants.  This forces me to agree with Alqatrkapa's assessment that the correct function to be differentiated is very likely
$f(t) = pe^t / (1-qe^t); \tag{3}$
and I refer the reader to the most excellent analysis in his answer for the derivative of this function.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!
