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Thread: Bases other than e... help?

  1. #1
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    Bases other than e... help?

    im stuck here, how do i do this?

    g(x) = ( log [10] x ) / (x^2)

    note:
    log [10] x-- this is log, sub 10, x... i didnt know how to write it out
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by runner07 View Post
    im stuck here, how do i do this?

    g(x) = ( log [10] x ) / (x^2)

    note:
    log [10] x-- this is log, sub 10, x... i didn't know how to write it out
    um, you didn't tell us what to do
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  3. #3
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    ahh sorry, it says to differentiate the function.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by runner07 View Post
    ahh sorry, it says to differentiate the function.
    ok. use the change of base formula for logarithms

    recall that: $\displaystyle \log_a b = \frac {\log_c b}{\log_c a}$

    so we can write: $\displaystyle \log_{10} x = \frac {\ln x}{\ln 10}$

    now continue
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    ok. use the change of base formula for logarithms

    recall that: $\displaystyle \log_a b = \frac {\log_c b}{\log_c a}$

    so we can write: $\displaystyle \log_{10} x = \frac {\ln x}{\ln 10}$

    now continue
    so then i would use the quotient rule? making it...

    (x^2)(lnx/ln10) - (lnx/ln10)(2x)
    (x^2)^2

    is that right?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by runner07 View Post
    so then i would use the quotient rule? making it...

    (x^2)(lnx/ln10) - (lnx/ln10)(2x)
    (x^2)^2

    is that right?
    no, that's not the quotient rule. try again
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  7. #7
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    oh wait, so would i get the derivative of lnx/ln10 first then plug it into that?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by runner07 View Post
    oh wait, so would i get the derivative of lnx/ln10 first then plug it into that?
    yes, remember the quotient rule: $\displaystyle \left( \frac {u}{v} \right)^{\prime} = \frac {u'v - v'u}{v^2}$

    also, it may be easier for you to write: $\displaystyle g(x) = \frac {1}{\ln 10} \cdot \frac {\ln x}{x^2}$
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  9. #9
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    Quote Originally Posted by Jhevon View Post
    yes, remember the quotient rule: $\displaystyle \left( \frac {u}{v} \right)^{\prime} = \frac {u'v - v'u}{v^2}$

    also, it may be easier for you to write: $\displaystyle g(x) = \frac {1}{\ln 10} \cdot \frac {\ln x}{x^2}$
    so first i would do the quotient rule for the lnx/x^2 right? and then the product rule with the 1/ln10 ?
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  10. #10
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    Quote Originally Posted by runner07 View Post
    so first i would do the quotient rule for the lnx/x^2 right? and then the product rule with the 1/ln10 ?
    Hello,

    not quite. Remember $\displaystyle \frac{1}{\ln(10)}$ is a constant.

    $\displaystyle g(x) = \frac {1}{\ln 10} \cdot \frac {\ln x}{x^2}$ will give:

    $\displaystyle g'(x) = \frac {1}{\ln 10} \cdot \frac {x^2 \cdot \frac{1}{x} - \ln (x) \cdot 2x}{x^4}$

    I'll leave the rest for you. (You can cancel out a factor - under certain conditions!)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by runner07 View Post
    so first i would do the quotient rule for the lnx/x^2 right? and then the product rule with the 1/ln10 ?
    as earboth already said, when we have a constant multiplier to a derivative, we can essentially forget about it and differentiate the (variable) function and then apply the constant multiplier when we're done. in fact, this is one of the properties of derivatives:

    $\displaystyle \frac {d}{dx} c \cdot f(x) = c \frac {d}{dx}f(x)$

    where $\displaystyle c$ is a constant
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  12. #12
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    Hello, runner07!

    You should have been shown the formula for "other bases".


    im stuck here, how do i differentiate this?
    . . $\displaystyle g(x) \:= \:\frac{\log_{10}x}{x^2}$
    You can always derive the formula . . . (Yeah, right!)


    We have: .$\displaystyle y \:=\:\log_b(x)$ . ... where $\displaystyle b$ is a positive real number $\displaystyle \neq 1$

    . . Re-write in exponential form: .$\displaystyle b^y \:=\:x$

    . . Take logs: .$\displaystyle \ln\left(b^y\right) \:=\:\ln(x)\quad\Rightarrow\quad y\!\cdot\!\ln(b) \:=\:\ln(x)$

    . . Differentiate implicitly: .$\displaystyle \frac{dy}{dx}\!\cdot\ln(b) \:=\:\frac{1}{x}$

    Therefore: . $\displaystyle \boxed{\frac{dy}{dx} \;=\;\frac{1}{\ln b}\!\cdot\!\frac{1}{x}} $


    So the derivative of $\displaystyle \log_{10}x$ is: .$\displaystyle \frac{1}{\ln10}\!\cdot\!\frac{1}{x}$


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  13. #13
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    You can use the product rule too

    $\displaystyle g(x)=\frac{\log_{10}x}{x^2}=\frac{\dfrac{\ln x}{\ln10}}{x^2}=\frac1{\ln10}\cdot\frac{\ln x}{x^2}$

    So, all we have to do it's take the derivative of $\displaystyle \frac{\ln x}{x^2}=x^{-2}\ln x$, now by the product rule

    $\displaystyle \ln 10\cdot g'(x)=-2x^{-3}\ln x+x^{-2}\cdot\frac1x=-\frac{2\ln x}{x^3}+\frac1{x^3}=\frac{1-2\ln x}{x^3}$
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