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Math Help - Functions

  1. #1
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    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.
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  2. #2
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    doesn't the logs turn into e
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  3. #3
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    Quote Originally Posted by theflyingcow View Post
    doesn't the logs turn into e
    That's a little bit vague. Can you be more specific?
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  4. #4
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    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)
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  5. #5
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    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.
    Last edited by badgerigar; November 3rd 2008 at 05:33 PM. Reason: dint notice Jameson's post
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  6. #6
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    i am sorry to say im still confuse. I stink when it comes to formulas. Can you show me step by step please
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  7. #7
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    badgerigar - Happens all the time. That's the curse of trying to make a length post.
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  8. #8
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    Please help I need to know this
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  9. #9
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    Quote Originally Posted by uniquereason81 View Post
    Please help I need to know this
    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.
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  10. #10
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    Sorry Jameson
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  11. #11
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    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?
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  12. #12
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    .
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  13. #13
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    Quote Originally Posted by theflyingcow View Post
    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
    Quote Originally Posted by uniquereason81 View Post
    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..
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  14. #14
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    what rule did you use
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  15. #15
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    Quote Originally Posted by uniquereason81 View Post
    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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