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Derivative PLEASE help

  1. Oct 18, 2009 #1
    Derivative... PLEASE help!!!

    Please compute M'(t) and M''(t)?

    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. jcsd
  3. Oct 19, 2009 #2

    lurflurf

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    Re: Derivative... PLEASE help!!!

    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?
     
  4. Oct 19, 2009 #3
    Re: Derivative... PLEASE help!!!

    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.
     
  5. Oct 19, 2009 #4

    lurflurf

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    Re: Derivative... PLEASE help!!!

    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
     
  6. Oct 19, 2009 #5

    arildno

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    Re: Derivative... PLEASE help!!!

    Well, we have:
    [tex]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}}[/tex]

    Then, we get:
    [tex]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})[/tex]

    Unless I've done something wrong, that is. :smile:
     
    Last edited: Oct 19, 2009
  7. Oct 19, 2009 #6

    arildno

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    Re: Derivative... PLEASE help!!!

    An alternative would be to simplify as follows:
    [tex]M(t)=\frac{(pe^{t})^{r}}{(1-qe^{t})^{r}}=(\frac{pe^{t}}{1-qe^{t}})^{r}=\frac{p^{r}}{(e^{-t}-q)^{r}}[/tex]

    You might use this as your basic expression prior to differentiation.
     
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