A lake is stocked with 500 fish and their population $t$ months after the fish were introduced is $\large p(t)=\frac{10,000}{1+19e^{-0.2t}}$. Observe that $p(0)=500$. At what rate is the fish population changing, one month after the lake is stocked?

Question

luigi0210 · Answer

@agent0smith @whpalmer4 @jigglypuff314

mathslover · Answer

You need to find $\cfrac{dP(t)}{dt} $ here. Right?

luigi0210 · Answer

Uh, I think so.

mathslover · Answer

We will solve this by using chain rule.

mathslover · Answer

Let us take, the denominator as "x" 

So, differentiation of $\cfrac{10^4}{x} = ? $

luigi0210 · Answer

$$\LARGE -\frac{10^4}{x^2}$$

mathslover · Answer

Right! And now we will differentiate the rest part. 

Differentiation of $1+ 19e^{-0.2t}$ = ?

luigi0210 · Answer

$$\LARGE (-0.2)(19e^{-0.2t})$$

mathslover · Answer

Right so you have : $\cfrac{-10^4}{(1+19e^{-0.2t})^2} \times (-0.2) \times (19e^{-0.2t})$

luigi0210 · Answer

What do we do from here? o.O

mathslover · Answer

Hmm, I think just simplifying this further is enough for us?

$\cfrac{3.8 \times 10^4 \times 19e^{-0.2t} }{ (1+19e^{-0.2t})^2}$

mathslover · Answer

$\cfrac{3.8 \times 10^4 \times e^{-0.2t} }{ (1+19e^{-0.2t})^2}$ 

- Sorry, I didn't remove 19 from there.