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Math Help - Differential equations

  1. #1
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    Differential equations

    Use the definition of the derivative to find f' (x).

    f (x)=  \frac {1}{\sqrt(x)}

    The answer I got is \frac{-\sqrt(x)}{2x}

    Can someone tell me if this is right? If it isn't, I will post my work to see where I went wrong.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yeloc View Post
    Use the definition of the derivative to find f' (x).

    f (x)=  \frac {1}{\sqrt(x)}

    The answer I got is \frac{-\sqrt(x)}{2x}

    Can someone tell me if this is right? If it isn't, I will post my work to see where I went wrong.
    it is wrong

    and this is not a differential equation
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    it is wrong

    and this is not a differential equation
    How is it not when you have to use a differential equation to solve it?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by yeloc View Post
    Use the definition of the derivative to find f' (x).

    f (x)=  \frac {1}{\sqrt(x)}

    The answer I got is \frac{-\sqrt{x}}{2x}

    Can someone tell me if this is right? If it isn't, I will post my work to see where I went wrong.
    If it was f'(x)=-\frac{\sqrt{x}}{2x^{\color{red}2}}, then you'd be correct!

    What's the derivative of x^{-\frac{1}{2}}?

    Recall that \frac{d}{\,dx}x^n=nx^{n-1}

    --Chris
    Last edited by Chris L T521; September 14th 2008 at 05:25 PM. Reason: ooppss... XD
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yeloc View Post
    How is it not when you have to use a differential equation to solve it?
    you are doing basic calculus. "differential equations" refers to something else. see here
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    If it was f'(x)=-\frac{\sqrt{x}}{2x^{\color{red}2}}, then you'd be correct!
    technically yes, technically no. it would depend on whether we are considering negative x's or positive or both. remember, \sqrt{x^2} = |x|

    it is better not to simplify in this case, so there is no ambiguities. just apply the chain rule, as you so rightly directed, and call it a day
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  7. #7
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    After I substituted x+ \Delta x in for x, and used the definition of the derivative of a function to set the rest of the problem up, how would I get  \Delta x out of the denominator?

    I used the conjugate method, then I got a common denominator. I just simplified from there on out. Is this not right?
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yeloc View Post
    After I substituted x+ \Delta x in for x, and used the definition of the derivative of a function to set the rest of the problem up, how would I get  \Delta x out of the denominator?

    I used the conjugate method, then I got a common denominator. I just simplified from there on out. Is this not right?
    you have the right idea. whether or not you did it right, i don't know
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