Question

How can i simplify this fraction. The task was to derive N(t) and then use simple algebra to simplify to a form such as N'(t) = #N - #N^2 where # are a mix of variables (a, b, c)

Answer 1

\[N(t) = \frac{ a }{ b+e^{-ct} }\]

Answer 2

I then derived it and got \[N'(t) = \frac{ -ace ^{-ct} }{ b^2 + 2be^{-ct} + e^{-2ct} }\]

Answer 3

"derive" is a wrong word
use "differentiate" instead

Answer 4

recall that the derivative of \(\dfrac{1}{f(t)}\) is \(-\dfrac{1}{f^2(t)}*f'(t)\)

Answer 5

\[N(t) = \frac{ a }{ b+e^{-ct} } \]

differentiate both sides with respect to \(t\) and get
\[N'(t) = -\frac{ a }{ (b+e^{-ct})^2 } *(b+e^{-ct})' = -\frac{ a }{ (b+e^{-ct})^2 } * (-ce^{-ct}) \]

Answer 6

with me so far ?

Answer 7

***\[ N'(t) = \frac{ -ace ^{-ct} }{ b^2 + 2be^{-ct} + e^{-2ct} }\] ***
that looks ok except I think it should be +ac up top.

also, you should keep the bottom in the form (a+b e^(-ct) )^2
because you want to get N out of that mess
in other words, write it like this:
\[ N' = \frac{a}{b+e^{-ct}} \cdot \frac{a}{b+e^{-ct}} \cdot \frac{c}{a} \cdot e^{-ct} \]

the first two expressions are N
to get rid of the e^-ct, use the original equation for N, and "solve for e-3t"
and plug that into the derivative and simplify.

Answer 8

@ganeshie8 yep, got that

Answer 9

see if you can follow the next steps :
\[N(t) = \frac{ a }{ b+e^{-ct} } \]

\[N'(t) = -\frac{ a }{ (b+e^{-ct})^2 } *(b+e^{-ct})' \\~\\~\\
= -\frac{ a }{ (b+e^{-ct})^2 } * (-ce^{-ct}) \\~\\~\\
= -\frac{ a }{ (b+e^{-ct})^2 } * (-ce^{-ct}) \\~\\
= -\frac{ a^2 }{ (b+e^{-ct})^2 } * \dfrac{(\color{red}
{b-b}-ce^{-ct})}{a} \\~\\~\\
= -N^2*(\dfrac{b}{a}-\dfrac{1}{N})\\~\\~\\
=N -\dfrac{b}{a}N^2
\]

Answer 10

wow. Thank you! I would have never thought to have done that.

Answer 11

you should note ganeshie8 made a mistake
\[ \frac{b+ce^{-ct}}{a} \]
is not 1/N (notice the extra factor of c), rather:
\[ \frac{1}{N}=\frac{b+e^{-ct}}{a} \]


once you reach
\[ N' = \frac{a}{b+e^{-ct}} \cdot \frac{a}{b+e^{-ct}} \cdot \frac{c}{a} \cdot e^{-ct} =N^2 \frac{c}{a} \cdot e^{-ct}\]

go back to the original equation
\[ N(t) = \frac{ a }{ b+e^{-ct} } \\ b+e^{-ct} = \frac{ a }{N} \\ e^{-ct} = \frac{ a }{N}-b
\]
and substitute for e^-ct :
\[ N' = N^2 \frac{c}{a} \cdot e^{-ct} \\
N' = N^2 \cdot \frac{c}{a} \cdot \left(\frac{ a }{N}-b \right)
\]

that simplifies to
\[ N'= cN - \frac{bc}{a} N^2 \]

Answer 12

i see your point. i missed the c as well that carried from deriving. thank you! i'm glad you both will get medals in the end. your help was greatly appreciated.