How did they derive at the derivative?

• Apr 6th 2011, 07:39 AM
scounged
How did they derive at the derivative?
I've just had a hard time with a problem in my maths book, and it ended up with me doing a step I think is totally unnecessary, but which I knew would give me the right answer. This is the problem:

In a boarding school the measles are spreading. The number of students N who fall ill follows the function

$\displaystyle \displaystyle{N(t)}=\frac{250}{1+249e^{-t}}$

where t is the number of days after the first student has fallen ill. Calculate and interpret

a) N'(7)
b) N''(7).

a) was quite easy, and after differenciating N(t) I got the derivative

$\displaystyle \displaystyle{N'(t)}=\frac{62250e^{-t}}{(1+249e^{-t})^{2}}$

which I then used to get that N'(7) is about 38.

Then I came to b). This is where I got problems.
I used the quotient rule and got this:

$\displaystyle \displaystyle{N''(t)}=\frac{-62250e^{-t}(1+249e^{-t})^{2}-2(62250e^{-t})(1+249e^{-t})(-249e^{-t})}{(1+249e^{-t})^{4}}$

however, this seems to be a faulty differenciation, at least when I put it in my calculator. I suspect that I've misused the chain rule in some way, but I don't know. I'm not very experienced with either the chain rule, the product rule, nor the quotient rule as I've just started counting with them.

The whole thing ended up with me getting frustrated and eventually expanding the denomenator (I think that this is an unnecessary step, and I'm not glad that I was forced to use it) in the N'(t) expression, and after differenciating, I got this:

$\displaystyle {N''(t)}=\displaystyle{\frac{3859562250e^{-3t}-62250e^{-t}}{(1+249e^{-t})^{4}}}$

which gave me the correct answer N''(7)= -23.7 , but looks rather clumsy in my opinion.

In the answers section of the book they had come up with another expression for N''(t), namely this:

$\displaystyle {N''(t)}=\displaystyle{\frac{31000500e^{-2t}-62250e^{-t}(1+249e^{-t})}{(1+249e^{-t})^{3}}}$

How did they do that?
• Apr 6th 2011, 08:59 AM
running-gag
$\displaystyle \displaystyle{N''(t)}=\frac{-62250e^{-t}(1+249e^{-t})^{2}-2(62250e^{-t})(1+249e^{-t})(-249e^{-t})}{(1+249e^{-t})^{4}}$

Simplification by $\displaystyle 1+249e^{-t}$ gives

$\displaystyle \displaystyle{N''(t)}=\frac{-62250e^{-t}(1+249e^{-t})-2(62250e^{-t})(-249e^{-t})}{(1+249e^{-t})^{3}}$

$\displaystyle {N''(t)}=\displaystyle{\frac{-62250e^{-t}(1+249e^{-t})+31000500e^{-2t}}{(1+249e^{-t})^{3}}}$
• Apr 6th 2011, 09:33 AM
scounged
Hmmm. I guess I was sloppy with my calculator then.