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Math Help - Where am i going wrong on taking the derivative?

  1. #1
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    Where am i going wrong on taking the derivative?

    The problem is \frac{t^8+9t+3}{t^2}


    I used the qoutent rule and got \frac{t^2(8t^7+9)-t^8+9t+3(2t)}{(t^2)^2}

    I ended with \frac{3t(2t^8+9t+2)}{t^4}

    This work isn't looking like the answer at all. Where am I going wrong?



    Thanks
    Last edited by Darkhrse99; August 5th 2009 at 08:59 AM.
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  2. #2
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    Quote Originally Posted by Darkhrse99 View Post
    The problem is \frac{t^8+9t=3}{t^2}


    I used the qoutent rule and got \frac{t^2(8t^7+9)-t^8+9t+3(2t)}{(t^2)^2}

    I ended with \frac{3t(2t^8+9t+2)}{t^4}

    This work isn't looking like the answer at all. Where am I going wrong?



    Thanks
    The = in the initial problem is supposed to be a +, I think? Anyway, when you did the Quotient Rule, you forgot to multiply the terms in the numerator (2nd part)....
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  3. #3
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    The middle work I left out is (8t^9+9t^2)-2T^9+18t^2+6t

    doing subtraction gives you \frac{6t^9+27t^2+6t}{(t^2)^2}

    I did the multiplication in the numerator.
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  4. #4
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    Quote Originally Posted by Darkhrse99 View Post
    The middle work I left out is (8t^9+9t^2)-2T^9+18t^2+6t

    doing subtraction gives you \frac{6t^9+27t^2+6t}{(t^2)^2}

    I did the multiplication in the numerator.
    I don't follow that... but what I'm saying is, the Quotient Rule gives

    \frac{t^2(8t^7+9)-(t^8+9t+3)(2t)}{t^4},

    which is not what you had.
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  5. #5
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    Quote Originally Posted by Darkhrse99 View Post
    \frac{t^8+9t+3}{t^2}
    I would break it up into three terms, which seems a lot easier to me.

    \frac{t^8}{t^2}+\frac{9t}{t^2}+\frac{3}{t^2}
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  6. #6
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    Quote Originally Posted by Jameson View Post
    I would break it up into three terms, which seems a lot easier to me.

    \frac{t^8}{t^2}+\frac{9t}{t^2}+\frac{3}{t^2}
    Jameson,

    Using the qoutent rule, I came up with \frac{-6t^9+3t^2+6t}{(t^2)^2}

    How can I simplify this more?
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  7. #7
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    Quote Originally Posted by AlephZero View Post
    I don't follow that... but what I'm saying is, the Quotient Rule gives

    \frac{t^2(8t^7+9)-(t^8+9t+3)(2t)}{t^4},

    which is not what you had.
    I had posted this problem the exact way you did in my first post.
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  8. #8
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    Quote Originally Posted by Darkhrse99 View Post
    I had posted this problem the exact way you did in my first post.
    When you multiply this out I get \frac{(8t^9+9t^2)-(2t^9+18t^2+6t)}{t^4}

    Is this correct?
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  9. #9
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    Quote Originally Posted by Darkhrse99 View Post
    The problem is \frac{t^8+9t+3}{t^2}


    I used the qoutent rule and got \frac{t^2(8t^7+9)-t^8+9t+3(2t)}{(t^2)^2}

    I ended with \frac{3t(2t^8+9t+2)}{t^4}

    This work isn't looking like the answer at all. Where am I going wrong?



    Thanks
    Quote Originally Posted by AlephZero View Post
    I don't follow that... but what I'm saying is, the Quotient Rule gives

    \frac{t^2(8t^7+9)-(t^8+9t+3)(2t)}{t^4},

    which is not what you had.
    Quote Originally Posted by Darkhrse99 View Post
    I had posted this problem the exact way you did in my first post.
    As you can see, you absolutely did not have it posted that way, and that is the beginning of where your calculations went wrong.
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  10. #10
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    Quote Originally Posted by AlephZero View Post
    As you can see, you absolutely did not have it posted that way, and that is the beginning of where your calculations went wrong.
    You right, I left the bracket's out by mistake and didn't sqaure the t^2

    Is my math correct though?
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  11. #11
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    Quote Originally Posted by Darkhrse99 View Post
    Jameson,

    Using the qoutent rule, I came up with \frac{-6t^9+3t^2+6t}{(t^2)^2}

    How can I simplify this more?
    Quote Originally Posted by Darkhrse99 View Post
    You right, I left the bracket's out by mistake and didn't sqaure the t^2

    Is my math correct though?
    If your answer is the one above, then I'd have to say no. If you've revised your answer, then you need to provide it, as I haven't seen you give another.
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  12. #12
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    Is the answer 6t^5+\frac{9}{t^2}+\frac{6}{t^3}?
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  13. #13
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    Quote Originally Posted by Darkhrse99 View Post
    The problem is \frac{t^8+9t+3}{t^2}
    work smart, not hard.

    \frac{t^8+9t+3}{t^2} = t^6 + 9t^{-1} + 3t^{-2}

    \frac{d}{dt}\left[t^6 + 9t^{-1} + 3t^{-2}\right] = 6t^5 - 9t^{-2} - 6t^{-3} = 6t^5 - \frac{9}{t^2} - \frac{6}{t^3}
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  14. #14
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    Quote Originally Posted by Darkhrse99 View Post
    Is the answer 6t^5+\frac{9}{t^2}+\frac{6}{t^3}?
    No. You have asked where your calculations went wrong, and I have shown you over and again where it is. As I said, you have not done the multiplication in the numerator correctly, because you've neglected to distribute the -1. So, once again, the numerator is

    t^2(8t^7+9)-(t^8+9t+3)(2t)

    =8t^9+9t^2-2t^9-18t^2-6t

    =6t^9-9t^2-6t.
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  15. #15
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    Quote Originally Posted by skeeter View Post
    work smart, not hard.

    \frac{t^8+9t+3}{t^2} = t^6 + 9t^{-1} + 3t^{-2}

    \frac{d}{dt}\left[t^6 + 9t^{-1} + 3t^{-2}\right] = 6t^5 - 9t^{-2} - 6t^{-3} = 6t^5 - \frac{9}{t^2} - \frac{6}{t^3}
    I believe you ment 6t^5-\frac{9}{t^2}-\frac{6}{t^3}?
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