Question

Derivative of population formula.
The formula is that of a butterfly population. It is y=6000/(1+46(0.6)^t)

Answer 1

I read the answer in the book but they don't explain the steps so I don't understand how they got the answer.

Answer 2

is it
\[\frac{6000}{1+46(.6)^t}\]

Answer 3

Yes

Answer 4

I changed its form for product rule applciation

Answer 5

ok the derivative of \(\frac{1}{f}\) is \(-\frac{f'}{f^2}\)

Answer 6

But this requires us to use chain rule as well and no matter how many times I do it mself I get the wrong answer

Answer 7

nah, product is too confusing
use
\[-\frac{f'}{f^2}\]

Answer 8

I amde it 6000(1+49(0.6)^t)^-1

Answer 9

I have never seen that what is that rule called

Answer 10

\[f(x)=1+46(.6)^t\]
\[f'(x)=46\ln(.6)(.6^t)\]

Answer 11

In the book they sue the product rule and the chain rule so ideally i wanna learn how to do it the way they did it

Answer 12

use*

Answer 13

you are making too much work for yourself

Answer 14

which rule?

Answer 15

f′(x)=46ln(.6)(.6t) <---- what rule is that! Where did you get the ln!

Answer 16

the derivative of \(b^x\) is \(\ln(b)\times b^x\) always

Answer 17

Can you explain to me in steps how you would do the whole things using product rule and chian rule step by step Ill follow on paper. Sorry for the hassle but I really need to internalize this process

Answer 18

yeah hold on one minute

Answer 19

I SUPER appreciate this btw

Answer 20

ok first of all we are not going to use the product rule or the quotient rule

Answer 21

even tho the book has their answer that way?

Answer 22

if i had
\[\frac{x-1}{x+1}\] i would use the quotient rule, because there is a variable in the numerator

Answer 23

but you have no variable in the numerator, just a number (constant)

Answer 24

right

Answer 25

if you want the derivative of
\[\frac{c}{f}\] where \(c\) is a constant, it is
\[-\frac{cf'}{f^2}\]

Answer 26

the quotient rule will give you the same thing, because the derivative of the numerator is zero if it is a constant

Answer 27

I guess the constant is kind of throwing me off

Answer 28

so rather than using
\[\left(\frac{f}{g}\right)'=\frac{gf'-fg'}{g^2}\] if \(f\) is a number \(c\) , you just get
\[-\frac{cg'}{g^2}\]

Answer 29

c=-f right

Answer 30

another way to look at this is as a chain rule problem

\[\frac{d}{dx}\frac{1}{x}=-\frac{1}{x^2}\] and so
\[\frac{d}{dx}\frac{1}{f(x)}=-\frac{1}{f^2(x)}\times f'(x)=-\frac{f'(x)}{f^2(x)}\]

Answer 31

in order to do your problem, the first thing you do is put parentheses around the denominator and square it

Answer 32

i don't mean actually square it, i just mean write it

Answer 33

1+49^2(0.6)^2t

Answer 34

\[f(t)=\frac{6000}{1+46(.6)^t}\]
\[f'(t)=\frac{-6000f'(t)}{(1+46\times .6^t)^2}\]

Answer 35

oh sorry
ok i got it

Answer 36

ok and then

Answer 37

bad algebra there in any case
you can't square like that....

Answer 38

ya i screwed up i noticed

Answer 39

the only think that is missing is \(f'(t)\)

Answer 40

by which i mean the derivative of the denominator
\[1+46\times .6^t\]

Answer 41

Bleh your explanation is great but i keep blanking out. I'm gonna review previous units I just took a long break and forgot a bunch of rules.

Answer 42

Thanks for the help, I just think I gotta drill derivative rules again

Answer 43

this is what you need to know to get that derivative
the derivative of 1 is 0
the derivative of \(b^x\) is \(b^x\ln(b)\)

Answer 44

so the derivative of your denominator is
\[46\ln(.6)\times .6^t\]

Answer 45

ya i understood that part thanks

Answer 46

ok then you are done
answer
\[\frac{-6000\times 46\ln(.6)\times .6^t}{(1+46\times .6^t)^2}\]

Answer 47

yw

Answer 48

thx buddy