Question

f(x)=log2 (x)^2/(x-1) and its bases other than e. i am really confused with this one, (x-2)/(ln2)(x)(x-1) is the answer and every time i find my answer its not right please help

Answer 1

This is really confusing, what is the question?

Answer 2

\[\Large\rm f(x)=\log_2\left(\frac{x^2}{x-1}\right)\]Is this the function..?

Answer 3

And what are you trying to do?
Derivative? Need more information.

Answer 4

yes, it was a mess to type sorry

Answer 5

find the derivative of the function, and with ln

Answer 6

Apply change of base before taking a derivative:\[\Large\rm \log_2(a)=\frac{\ln(a)}{\ln(2)}\]

Answer 7

\[\Large\rm f(x)=\frac{\ln\left(\frac{x^2}{x-1}\right)}{\ln(2)}\]Understand what I did there? :o

Answer 8

i see the ln 2 but you would put the function on the top?

Answer 9

\[\Large\rm \log_b(\color{royalblue}{c})=\frac{\ln(\color{royalblue}{c})}{\ln(b)}\]
Applying it gives us:\[\Large\rm \log_2\left(\color{royalblue}{\frac{x^2}{x-1}}\right)=\frac{\ln\left(\color{royalblue}{\frac{x^2}{x-1}}\right)}{\ln(2)}\]

Answer 10

I'm not exactly sure what you were asking.
Hopefully the color coding helps a little bit.

Answer 11

Yes, this is all pre-work before we take a derivative.

Answer 12

Now pull the denominator out of the way since it's just a constant,\[\Large\rm f(x)=\frac{1}{\ln 2}\cdot\ln\left(\frac{x^2}{x-1}\right)\]

Answer 13

So from here, take the derivative the way you remember.
Apply your rule for derivative of natural log, then chain rule into quotient rule.

Answer 14

No good? =O
Too confusing?

Answer 15

well i see how to use the quotient rule but the chain rule?

Answer 16

Yah maybe that was sloppy the way I said it.
Chain rule is just telling you that you need to multiply by the derivative of the inner function. So chain rule is saying -> Hey after you take derivative of the natural log, apply quotient rule to the inside.

Maybe I shouldn't had listed it like a "step".
It's just a command telling you to apply quotient rule next.

Answer 17

\[\Large\rm f'(x)=\frac{1}{\ln 2}\cdot\frac{1}{\left(\frac{x^2}{x-1}\right)}\color{royalblue}{\left(\frac{x^2}{x-1}\right)'}\]Chain rule is what makes this blue part show up^
The way you actually DEAL WITH IT, is quotient rule, as you indicated.

Answer 18

i get with the quotient rule \[x^{2}-1/(x-1)^{2} that correct?\]

Answer 19

For just the quotient rule portion?
Hmm I'm coming up with:\[\Large\rm \frac{x^2-2x}{(x-1)^2}\]Hmm let's see what went wrong...
This will let me double check my own work as well.

Answer 20

\[\Large\rm \color{royalblue}{\left(\frac{x^2}{x-1}\right)'}=\frac{\color{royalblue}{(x^2)'}(x-1)-x^2\color{royalblue}{(x-1)'}}{(x-1)^2}\]
\[\Large\rm =\frac{\color{orangered}{(2x)}(x-1)-x^2\color{orangered}{(1)}}{(x-1)^2}\]

Answer 21

\[\Large\rm =\frac{2x^2-2x-x^2}{(x-1)^2}\]

Answer 22

I think I did that correctly.
Did you forget to distribute the 2x maybe?

Answer 23

yeah, your right

Answer 24

\[\Large\rm f'(x)=\frac{1}{\ln 2}\cdot\frac{1}{\left(\frac{x^2}{x-1}\right)}\cdot\color{royalblue}{\left(\frac{x^2}{x-1}\right)'}\]So I guess we end up with:\[\Large\rm f'(x)=\frac{1}{\ln 2}\cdot\frac{1}{\left(\frac{x^2}{x-1}\right)}\cdot\color{orangered}{\frac{2x^2-2x-x^2}{(x-1)^2}}\]

Answer 25

Which can be simplified a little bit further.

Answer 26

yeah i was just abou to say 2x^2-x^2 can be X^2

Answer 27

\[\Large\rm f'(x)=\frac{1}{\ln 2}\cdot\frac{1}{\left(\frac{x^2}{x-1}\right)}\cdot\frac{x^2-2x}{(x-1)^2}\]
How bout that middle part?
Remember what to do when you divide by a fraction?

Answer 28

do the reciprocal

Answer 29

\[\Large\rm f'(x)=\frac{1}{\ln 2}\cdot\frac{x-1}{x^2}\cdot\frac{x^2-2x}{(x-1)^2}\]Ok good.
That should make canceling things out much easier.

Answer 30

In the last piece, factor an x out of each term in the numerator,\[\Large\rm f'(x)=\frac{1}{\ln 2}\cdot\frac{x-1}{x^2}\cdot\frac{x(x-2)}{(x-1)^2}\]You should see the dominoes all tipper over at this point c:

Answer 31

See some stuff you can cross out?

Answer 32

(x-1) correct?

Answer 33

\[\Large\rm f'(x)=\frac{1}{\ln 2}\cdot\frac{\cancel{x-1}}{x^2}\cdot\frac{x(x-2)}{(x-1)^{\cancel{2}1}}\]Yah, there's something else though.

Answer 34

the x's too

Answer 35

\[\Large\rm f'(x)=\frac{1}{\ln 2}\cdot\frac{1}{x^{\cancel{2}1}}\cdot\frac{\cancel{x}(x-2)}{(x-1)}\]Ah yes good.

Answer 36

Now just combine all the pieces and it should match the answer which was given.

Answer 37

Kind of a tough problem.
No only do you have a few derivative steps to work through,
but you also had to remember your log rule, and then a ton of algebra steps to simplify it down.

Answer 38

yeah, thanks zepdrix, that help :)

Answer 39

cool c: