1. Oct 18, 2009

### Taylor1234

M(t) =

(pe^t)^r
------------
(1-qe^t)^r

Can use R(t) = ln M(t) if you choose to.

Any help is greatly appreciated!!!

2. Oct 19, 2009

### lurflurf

Are p,q, and r constants or functions of t?
Do you know the derivative of e^t?
The chain rule?
The power rule?
The quotient rule?

3. Oct 19, 2009

### Taylor1234

p, q, and r are constants. Yes, I know that the derivative of e^t is e^t and all of the derivative rules, but after performing the quotient rule one time I got a huge mess that I could not seem to condense down. That is what I need help with.

Ultimately, I am trying to come up with M'(t) = r(1/p) but I cannot seem to arrive at that conclusion on my own.

4. Oct 19, 2009

### lurflurf

Try to simplify M before differentiation.
It is not possible that M'(t) = r(1/p), if that were so M(t) would have to be r(1/p)t.
If you know about implicit differentiation you could differentiate
M(t)*(1-qe^t)^r=(pe^t)^r

5. Oct 19, 2009

### arildno

Well, we have:
$$M'(t)=\frac{r*(pe^{t})^{r-1}*(pe^{t})}{(1-qe^{t})^{r}}-\frac{(pe^{t})^{r}}{ (1-qe^{t})^{r}}*\frac{r(1-qe^{t})^{r-1}*(-qe^{t})}{(1-qe^{t})^{r}}=rM(t)+M(t)*\frac{rqe^{t}}{1-qe^{t}}=\frac{rM(t)}{1-qe^{t}}$$

Then, we get:
$$M''(t)=\frac{rM'(t)}{1-qe^{t}}+\frac{rM(t)qe^{t}}{(1-qe^{t})^{2}}=\frac{r^{2}M(t)}{(1-qe^{t})^{2}}+\frac{rM(t)qe^{t}}{(1-qe^{t})^{2}}=\frac{rM(t)}{(1-qe^{t})^{2}}(r+qe^{t})$$

Unless I've done something wrong, that is.

Last edited: Oct 19, 2009
6. Oct 19, 2009

### arildno

$$M(t)=\frac{(pe^{t})^{r}}{(1-qe^{t})^{r}}=(\frac{pe^{t}}{1-qe^{t}})^{r}=\frac{p^{r}}{(e^{-t}-q)^{r}}$$