1. ## Functions

These are the last three problems that I need to know before the test .
Find f(x)'

A) f(x) = log (1-x)^2

B) f(x) = log9(2^x + 1)

C) f(x) = 2^(log(3)x)

The 9 is subscript in problem B
The 3 is subscript in problem C
Thanks alot for the help.

2. doesn't the logs turn into e

3. Originally Posted by theflyingcow
doesn't the logs turn into e
That's a little bit vague. Can you be more specific?

4. You definitely need to know how to find the derivative of $\log_{a}x$. Using a change of base this becomes $\frac{\ln(x)}{\ln(a)}$ and now this can be differentiated to get $\frac{1}{x\ln(a)}$.

For the first two, they aren't that bad just use the chain rule. The third one you can use that if $f=a^x$, then $f'=a^x \ln(a)$

5. Edit: whoops, sorry Jameson, didn't see that you had posted again

These are the last three problems that I need to know before the test .
Find f(x)'
I haven't seen this notation before, and I am assuming that f(x)' is the derivative of f(x). If this is not the case, please ignore me.

A) f(x) = log (1-x)^2
Remember the formula $\frac{d}{dx}\log(g(x)) = \frac{g'(x)}{g(x)}$ and/or use the chain rule.

B) f(x) = log9(2^x + 1)
Use $\log_a(b) = \frac{\log(b)}{\log(a)}$ and $a^x = e^{\log(a)x}$

C) f(x) = 2^(log(3)x)
Use all the formulas from the other 2 questions.

6. i am sorry to say im still confuse. I stink when it comes to formulas. Can you show me step by step please

7. badgerigar - Happens all the time. That's the curse of trying to make a length post.

9. Originally Posted by uniquereason81
This kind of post is known as "begging". This is your warning. Do not make these kind of posts. I know you need help and so does anyone reading this thread. It will get answered, just be a little patient.

10. Sorry Jameson

11. a) $f=(\log (1-x))^2$

Now like I told you, convert the log base 10 to log base "e".

$f= \left( \frac{\ln(1-x)}{\ln(10)} \right) ^2$

$f' = 2 \left( \frac{\ln(1-x)}{\ln(10)} \right) * \left( \frac{-1}{(1-x)\ln(10)} \right)$

Now I fear after all this that your problem was actually $f=\log[(1-x)^2]$. Which one is it?

12. .

13. Originally Posted by theflyingcow
doesn't the logs turn into e
no..? isnt the rule of logs : f(x)=log a (f(x)) , where a is a subscript. therefore f ' (x)= 1 1
___ . ___ . f ' (x)
ln a f(x)

sorry if you dont understand what i wrote..

so basically, for
Originally Posted by uniquereason81
A) f(x) = log (1-x)^2
what you do is :
f ' (x) = 1 over (ln 10) times 1 over (1-x)^2 times 2(1-x)times(-1)

hope this helps..

14. what rule did you use

15. Originally Posted by uniquereason81
what rule did you use
the log rule. it looks a little messed up in my previous post (sorry).
basically what it says is (in words): (1 over ln a) times (1 over f(x)) times f'(x).
a being the subscript.

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