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Math Help - The derivative of 1/x

  1. #1
    Senior Member Paze's Avatar
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    The derivative of 1/x

    Hi.

    I'm having problems understanding \frac{dy}{dx}\(\frac{1}{x})

    I always get 0 when I try to solve it with conventional algebra but the answer is \frac{-1}{x^2}

    Bonus question: How do I get the enclosing brackets in my tex code to cover \frac{1}{x} completely, instead of being small? Also, how do I do a space between \frac{dy}{dx} and \frac{1}{x} ?

    Thanks!
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  2. #2
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    Re: The derivative of 1/x

    Quote Originally Posted by Paze View Post
    I'm having problems understanding \frac{dy}{dx}\(\frac{1}{x})
    I always get 0 when I try to solve it with conventional algebra but the answer is \frac{-1}{x^2}
    Bonus question: How do I get the enclosing brackets in my tex code to cover \frac{1}{x} completely, instead of being small? Also, how do I do a space between \frac{dy}{dx} and \frac{1}{x} ?
    [tex]\frac{dy}{dx}\left(\frac{1}{x}\right)[/tex] gives \frac{dy}{dx}\left(\frac{1}{x}\right)

    \frac{dy}{dx}\left(\frac{1}{x}\right)=\frac{dy}{dx  }(x^{-1})

    -x^{-1-1}=x^{-2}=\frac{-1}{x^2}
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  3. #3
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    Re: The derivative of 1/x

    \frac{1}{x}=x^{-1}

    \frac{dy}{dx}(x^{-1})=-1*x^{-2}=\frac{-1}{x^2}
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: The derivative of 1/x

    Not sure what you mean by trying to solve with conventional algebra. Using the definition of the derivative you have:

    \frac {dy}{dx} = \lim_{h \to 0} \frac {f(x+h)-f(x)} h

     \frac {d(\frac 1 x)}{dx} = \lim _{h \to 0} \frac {\frac 1 {x+h} - \frac 1 x} h = \lim _{h \to 0} \frac {x - (x+h)}{x(x+h) h} = \lim _{h \to 0} \frac {-1} {x(x+h)} = \frac {-1} {x^2}

    Bonus questions: to make the parentheses larger you can use "\left(" and "\right)", like this:
     \left( \frac {-1} {x^2} \right)

    To make a space use "\ " - that's backslash followed by a space: "a b c" yields  a b c, wheras "a \ b \ c" yields  a \ b \ c.
    Last edited by ebaines; January 29th 2013 at 11:08 AM.
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  5. #5
    Senior Member Paze's Avatar
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    Re: The derivative of 1/x

    Thanks guys. I see how it works with this rule, but can you put it up in the fundamental formula: \frac{f(x)\frac{1}{x+\triangle x}-f(x)}{\triangle x}

    Okay I have no idea how this tex code is gonna come out seeing as I can't preview my answer, but I am asking to see the proof in the fundamental formula: (f(x+delta X) - f(x))/delta x

    Edit: holy sh** that came out almost right..
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  6. #6
    MHF Contributor ebaines's Avatar
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    Re: The derivative of 1/x

    Quote Originally Posted by Paze View Post
    Thanks guys. I see how it works with this rule, but can you put it up in the fundamental formula: \frac{f(x)\frac{1}{x+\triangle x}-f(x)}{\triangle x}
    Did you see my reply in post #4 of this thread?

    Quote Originally Posted by Paze View Post
    Okay I have no idea how this tex code is gonna come out seeing as I can't preview my answer, but I am asking to see the proof in the fundamental formula: (f(x+delta X) - f(x))/delta x
    You should have a "Go Advanced" button to the lower right, below the box where you type your reply - click it and you'll see a preview of your post.
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  7. #7
    Senior Member Paze's Avatar
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    Re: The derivative of 1/x

    Quote Originally Posted by ebaines View Post
    Did you see my reply in post #4 of this thread?



    You should have a "Go Advanced" button to the lower right, below the box where you type your reply - click it and you'll see a preview of your post.
    I noticed it now!
    Thank you and thank you all for your help!
    Last edited by Paze; January 29th 2013 at 10:44 AM.
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  8. #8
    Senior Member Paze's Avatar
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    Re: The derivative of 1/x

    Hold on...I'm trying to get this to work.. I can't seem to get it to work!

    In this: http://latex.codecogs.com/png.latex?%20\frac%20{d(\frac%201%20x)}{dx}%20=%20 \lim%20_{h%20\to%200}%20\frac%20{\frac%201%20{x+h} %20-%20\frac%201%20x}%20h%20=%20\lim%20_{h%20\to%200}% 20\frac%20{x%20-%20(x+h)}{x(x+h)%20h}%20=%20\lim%20_{h%20\to%200}% 20\frac%20{-1}%20{x(x+h)}%20=%20\frac%20{-1}%20{x^2}

    (sorry for long link)

    How does the swap between numerators happen and how does x-x-h become -1? -h becomes 0??

    Thanks.
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  9. #9
    Senior Member Paze's Avatar
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    Re: The derivative of 1/x

    Quote Originally Posted by ebaines View Post
    Not sure what you mean by trying to solve with conventional algebra. Using the definition of the derivative you have:


     \frac {d(\frac 1 x)}{dx} = \lim _{h \to 0} \frac {\frac 1 {x+h} - \frac 1 x} h = \lim _{h \to 0} \frac {x - (x+h)}{x(x+h) h} = \lim _{h \to 0} \frac {-1} {x(x+h)} = \frac {-1} {x^2}
    I mean in this..Link doesn't seem to work. So sorry for the confusion.
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  10. #10
    MHF Contributor ebaines's Avatar
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    Re: The derivative of 1/x

    When you have an expression of the form  \frac 1 a - \frac 1 b it may be useful to multiply the  \frac 1 a term by  (\frac b b ) and the  \frac 1 b term by  ( \frac a a ) :

     \frac 1 a - \frac 1 b = \frac {b}{b} ( \frac 1 a) - \frac a a (\frac 1 b) = \frac {b-a}{ab}

    The detailed steps are:

     \lim _{h \to 0} \frac {\frac 1 {x+h} - \frac 1 x} h

    = \lim _{h \to 0} \frac {\frac {x}{x(x+h)} - \frac {x+h}{x(x+h)}}h

    = \lim _{h \to 0} \frac {x - (x+h)}{x(x+h) h} = \lim _{h \to 0} \frac {-h}{x(x+h)h} = \lim _{h \to 0} \frac {-1} {x(x+h)} = \frac {-1} {x^2}

    Hope this helps.
    Last edited by ebaines; January 29th 2013 at 11:49 AM.
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  11. #11
    Senior Member Paze's Avatar
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    Re: The derivative of 1/x

    Quote Originally Posted by ebaines View Post
    When you have an expression of the form  \frac 1 a - \frac 1 b it may be useful to multiply the  \frac 1 a term by  (\frac b b ) and the  \frac 1 b term by  ( \frac a a ) :

     \frac 1 a - \frac 1 b = \frac {b}{b} ( \frac 1 a) - \frac a a (\frac 1 b) = \frac {b-a}{ab}

    The detailed steps are:

     \lim _{h \to 0} \frac {\frac 1 {x+h} - \frac 1 x} h

    = \lim _{h \to 0} \frac {\frac {x}{x(x+h)} - \frac {x+h}{x(x+h)}}h

    = \lim _{h \to 0} \frac {x - (x+h)}{x(x+h) h} = \lim _{h \to 0} \frac {-h}{x(x+h)h} = \lim _{h \to 0} \frac {-1} {x(x+h)} = \frac {-1} {x^2}

    Hope this helps.
    That clears it up! Thank you very much.
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