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Math Help - Finding Derivatives

  1. #1
    Member VitaX's Avatar
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    Finding Derivatives

    Find the derivative:

    \frac{dv}{dt} if v=t-\frac{1}{t}

    My answer is -\frac{2}{t}

    I can show work later if needed if this is wrong. Just need to be sure I did this one right before I go on and do others.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    dv/dt = 1 + 1/t^2

    since dt/dt = 1 and d(1/t)/dt = -1/t^2

    how did you get -2/t ?
    Last edited by Calculus26; September 23rd 2009 at 06:49 PM.
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  3. #3
    Member VitaX's Avatar
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    By definition \frac{f(x+h) - f(x)}{h} as h goes to 0.I first changed t - \frac{1}{t} = \frac{t^2 - 1}{t}. Plugged that in and got \frac{1}{h}[(\frac{(t+h)^2 - 1}{t+h} - (\frac{t^2 - 1}{t})]. Common denominator is t(t + h). Then it becomes \frac{1}{h}[\frac{t^2 + 2ht + h^2 - 1 - t^2 - 1}{t(t + h)}]. Then \frac{1}{h}[\frac{h^2 + 2ht - 2}{t(t + h)}]. Then \frac{h^2 + 2ht - 2}{ht(t + h)}. Then \frac{h + 2 - 2}{t + h} Substitute 0 in for h and you are left with -\frac{2}{t}

    Edit: Looks like 2 crosses out in the numerator so you are left with 0 ... hows this?
    Last edited by VitaX; September 23rd 2009 at 07:02 PM.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    See attachment it is easier if you don't combine

    By the way in my original post I ha a neg sign wrong but have corrected it
    Attached Thumbnails Attached Thumbnails Finding Derivatives-limit.jpg  
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  5. #5
    MHF Contributor Calculus26's Avatar
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    By the way if you do combine you have what is in the following attachment
    Attached Thumbnails Attached Thumbnails Finding Derivatives-limit2.jpg  
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  6. #6
    Member VitaX's Avatar
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    When I click the attachments my screen just goes dark. Nothing appears. Maybe an error on my pc in loading it. But I found a mistake in my work and am left with \frac{0}{t} which is just 0. Hows that?
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  7. #7
    MHF Contributor Calculus26's Avatar
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    right click on the attachment and click on open link
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  8. #8
    MHF Contributor Calculus26's Avatar
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    of course I meant click on "open link"
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  9. #9
    Member VitaX's Avatar
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    So where did I go wrong in my post above?

    Edit I think I made a sign error moving all under one denominator. So I think the answer is \frac{2}{t} if not then I'm completely lost as I did this problem the same basic way as my teacher showed us in class.
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  10. #10
    MHF Contributor Calculus26's Avatar
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    See my second attachment--It is more helpful for you if you find your own

    mistakes
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  11. #11
    Member VitaX's Avatar
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    Scratched this post. I understand everything you did and my mistakes. Thanks for your help.
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  12. #12
    MHF Contributor Calculus26's Avatar
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    Let h = 0

    you get (t^2 + 1)/(t^2) = 1 + 1/t^2
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  13. #13
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    you are trying to cancel terms that you can't cancel. since the top is a sum and not a product, you can't cancel out the "t's".. You will get (t^2+1)/t^2, which is the same as writing

    (t^2/t^2)+ 1/t^2
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  14. #14
    MHF Contributor Calculus26's Avatar
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    you are trying to cancel terms that you can't cancel. since the top is a sum and not a product, you can't cancel out the "t's".. You will get (t^2+1)/t^2, which is the same as writing

    (t^2/t^2)+ 1/t^2


    which simplifies to?
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  15. #15
    Member VitaX's Avatar
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    Yah I see it thats why I deleted my post above as it was dumb of me not to see it.
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