Question

Derivative of a log in the denominator.

Answer 1

\[\frac{ 1 }{ \log _{2}x }\]

Answer 2

do you know change of base formula?

Answer 3

\(\frac{d}{dx}\frac{1}{f}=-\frac{f'}{f^2}\)

Answer 4

Yikes. I learned it last year, but I don't remember it :/

Answer 5

in your case, \(f(x)=\log_2(x), f'(x)=\frac{1}{x\ln(2)}\)

Answer 6

guess you can use change of base as well right?
\[\log_2(x)=\frac{\ln(x)}{\ln(2)}\]

Answer 7

making your function
\[\frac{1}{\log_2(x)}=\frac{\ln(2)}{\ln(x)}\] you will get the same thing when you take the derivative, and don't forget that \(\ln(2)\) is just a constant, so DO NOT go and use the quotient rule

Answer 8

And then the derivative of ln(x) is 1/x? Or do I need the chain rule for that?

Answer 9

yes, but don't forget your log is in the denominator

Answer 10

So then the x is moved to the numerator, yes?

Answer 11

when i said do not use the quotient rule i meant use this
\[\left(\frac{1}{f}\right)'=\frac{-f'}{f^2}\]

Answer 12

how do you want to write your function? as
\[f(x)=\frac{1}{\log_2(x)}\] or as \[f(x)=\frac{\ln(2)}{\ln(x)}\]?
maybe we should just stick with the first one

Answer 13

for the first one, you just need the derivative of \(\log_2(x)\) which is \(\frac{1}{x\ln(2)}\)
put that in the numerators, put \((\log_2(x))^2\) in the denominator, and put a minus sign in front of the whole thing

Answer 14

you get
\[-\frac{\frac{1}{x\ln(2)}}{(\log_2(x))^2}\] and you can simplify with some algebra to get
\[-\frac{1}{(\log_2(x))^2x\ln(2)}\]

Answer 15

\[\frac{ 300 }{ 1+2^{4-t} }\]

Answer 16

i take it this is another question right?

Answer 17

This equation models the spread of a rumor at a school where t is the number of days after the rumor first started to spread.
And yes! It is :)

Answer 18

they really stretch for these word problems don't they "model for a rumor" indeed

Answer 19

what are you supposed to answer?

Answer 20

i means does it just ask for the derivative, or is there some part to the question

Answer 21

Estimate the initial number of students who first heard the rumor. An.d then they want how fast the rumor was spreading after 4 days

Answer 22

initial number means when you start counting, so replace \(t\) by \(0\)

Answer 23

then take the derivative, evaluate at \(t=4\)

Answer 24

For the first part, I just plug in 1 for t right? And then for the second part, I think I need the derivative?

Answer 25

no for the first part you plug in 0 for t, not 1

Answer 26

again for the derivative, the denominator will be \((1+2^{4-t})^2\)
the numerator will be \(300\times 2^{4-t}\times \ln(2)\)

Answer 27

Why thought? Why do you square the denominator and multiply it on top too?

Answer 28

the derivative of \(1+2^{4-t}\) is \(2^{4-t}\ln(2)\times (-1)\) by the chain rule
that \(-1\) is what makes the numerator positive

Answer 29

you square the denominator, multiply by minus the derivative of the denominator on top

Answer 30

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

in your case \(f(x)=1+2^{4-t},f'(t)=-2^{4-t}\ln(2)\)

Answer 31

so your "final answer" for the derivative works out to
\[\frac{300\times \ln(2)\times 2^{4-t}}{(1+2^{4-t})^2}\]

Answer 32

replacing \(t\) by 4 gives an easy calculation since both exponents are zero

Answer 33

Wait...I'm sorry for not getting this after you explained it so nicely. Can you show me one more time how/why you are getting \[-2^{4-t} \ln2\]? You said you used the chain rule for this?

Answer 34

sure

Answer 35

but first it is good to know that the derivative of \(b^x\) is \(b^x\ln(b)\)
you good with that one ?

Answer 36

OH MY GOD. I completely forgot that rule....WOW. That explains why you were getting that answer!! Argh. I'm so sorry. Yes, I understand what you are doing now.

Answer 37

ok good, then it should be easy right? the derivative of \(2^t\) is \(2^t\ln(2\) but the derivative of \(2^{4-t}\) is \(2^{4-t}\ln(2)\times (-1)\) because we have to take the derivative of \(4-t\) by the chain rule

Answer 38

that is why the numerator of your derivative is
\[300\ln(2)2^{4-t}\] because of the minus sign in the rule for reciprocals and the other minus sign from the chain rule above

Answer 39

your last job is to take
\[\frac{300\times \ln(2)\times 2^{4-t}}{(1+2^{4-t})^2}\] and replace \(t\) by \(4\)
you get
\[\frac{300\ln(2)}{4}\]

Answer 40

You are magnificent. You've taught me so much in the few minutes that we've been exchanging comments than my teacher taught me in an entire class period.

Answer 41

thank you for the compliment, glad to help
(blush)