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Math Help - Quotient rule

  1. #1
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    Quotient rule

    I need help.

    I want to use the quotient rule to find \frac{d}{dx}[\frac{3t^1^6-4}{t^6}]

    So I do
    \frac{t^6(\frac{d}{dx}[3t^1^6-4]) - (3t^1^6-4)(\frac{d}{dx}[t^6])}{(t^6)^2}

    = \frac{t^6(48t^1^5) - (3t^1^6-4)(6t^5)}{t^1^2}
    =  \frac{48t^2^1-18t^2^1-24t^5}{t^1^2}
    =  \frac{30^2^1-24t^5}{t^1^2}

    = 30^9-24t^-^7

    but the answer in the book is
    = 30^9+24t^-^7

    What Am I doing wrong?
    Last edited by delgeezee; October 6th 2011 at 10:56 AM.
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  2. #2
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    Re: Quotient rule

    Quote Originally Posted by delgeezee View Post
    I need help.

    I want to use the quotient rule to find \frac{d}{dx}[\frac{3t^1^6-4}{t^6}]

    So I do
    \frac{t^6(\frac{d}{dx}[3t^1^6-4]) - (3t^1^6-4)(\frac{d}{dx}[t^6])}{(t^6)^2}

    = \frac{t^6(48t^1^5) - (3t^1^6-4)(6t^5)}{t^1^2}
    =  \frac{48t^2^1-18t^2^1-24t^5}{t^1^2}
    =  \frac{30^2^1-24t^5}{t^1^2}

    = 30^9-24t^-^7

    but the answer in the book is
    = 30^9+24t^-^7

    What Am I doing wrong?
    Do you have to use the quotient rule? It is far easier to say that
    f(t) = \dfrac{3t^{16}-4}{t^6} = \dfrac{3t^{16}}{t^6} - \dfrac{4}{t^6} = 3t^{10} - 4t^{-6}

    Now you can use the power rule to find f'(t)


    Your mistake is when you multiplied out the second term in the numerator: -(3t^{16}-4)(6t^5) = =18t^{21} + 24t^5

    The minus sign becomes a plus because you're subtracting a negative -(-24t^5)
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    Re: Quotient rule

    I would like to be able to find the solution both ways so that I can double check my finals answers. It gives me a degree of self-assurance. But the alternative way you showed is indeed easier.
    Quote Originally Posted by e^(i*pi) View Post
    Do you have to use the quotient rule? It is far easier to say that
    f(t) = \dfrac{3t^{16}-4}{t^6} = \dfrac{3t^{16}}{t^6} - \dfrac{4}{t^6} = 3t^{10} - 4t^{-6}

    Now you can use the power rule to find f'(t)




    Your mistake is when you multiplied out the second term in the numerator: -(3t^{16}-4)(6t^5) = =18t^{21} + 24t^5

    The minus sign becomes a plus because you're subtracting a negative -(-24t^5)
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  4. #4
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    e^(i*pi)'s Avatar
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    Re: Quotient rule

    Quote Originally Posted by delgeezee View Post
    I need help.

    I want to use the quotient rule to find \frac{d}{dx}[\frac{3t^1^6-4}{t^6}]

    So I do
    \frac{t^6(\frac{d}{dx}[3t^1^6-4]) - (3t^1^6-4)(\frac{d}{dx}[t^6])}{(t^6)^2}

    = \frac{t^6(48t^1^5) - (3t^1^6-4)(6t^5)}{t^1^2}
    No problem.

    =  \frac{48t^{21}-(18t^{21}-24t^5)}{t^{12}}
    Here is where you went wrong, I have put brackets round the second term in the numerator to make it easier to see. What you did is say that -(18t^{21}-24t^5) = -18t^{21} - 24t^5 but the minus sign needs to be applied to both terms. I find it helps to imagine the minus sign as -1 and to then use the distributive law:

    -1(18t^{21} - 24t^5) = (-1 \cdot 18t^{21}) + (-1 \cdot -24t^5) = -18t^{21} + 24t^5

    I may have explained this poorly so please post back if you don't understand (and it's good to check both ways)
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