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Math Help - Need some help on differentiation

  1. #1
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    Need some help on differentiation

    Differentiate:

    1. F(y)=(\frac{1}{y^2}-\frac{3}{y^4})(y+5y^3)

    2. y=\frac{t^2}{3t^2-2t+1}

    3. y=\frac{v^3-2v\sqrt{v}}{v}

    I tried these and got 2 pages worth of arithmetic for one problem so I know I'm doing something wrong here.
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  2. #2
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    okay, i figured 2 out:

    \frac{t^2}{3t^2-2t+1)}=
    \frac{(3t^2-2t+1)\frac{d}{dx}(t^2)-(t^2)\frac{d}{dx}(3t^2-2t+1)}{(3t^2-2t+1)^2}=
    \frac{(3t^2-2t+1)(2t)-(t^2)(6t-2)}{(3t^2-2t+1)^2}=
    \frac{2t(1-t)}{(3t^2-2t+1)^2}
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  3. #3
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    Here is my thoughts on #1:

    1. <br />
F(y)=(\frac{1}{y^2}-\frac{3}{y^4})(y+5y^3)<br />

    I first do some algebra with it,

    Step#1 \frac {y+5y^3}{y^2} - \frac {3y+15y^3}{y^4}

    Step#2 \frac {y}{y^2} + \frac {5y^3}{y^2} - \frac  {3y}{y^4} - \frac  {15y^3}{y^4}

    Step#3 \frac {1}{y} + 5y - \frac {3}{y^3} - \frac {15}{y}

    Step#4 5y - 3y^{-3} - 14y^{-1}

    Now you should be able to solve it, I hope I did this right.

    KK
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  4. #4
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    Problem 3:

    Same thing, do the algebra first.

    <br />
y=\frac{v^3-2v\sqrt{v}}{v}<br />

    Step#1  \frac {v^3}{v} - \frac {2v\sqrt{v}}{v}

    Step#2  v^2 - 2v^{1/2}

    Now it is just straight forward differenation.

    KK
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  5. #5
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    i think your partial fraction decompostion for question 2 is wrong.. please check..
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  6. #6
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    Hello, c_323_h!

    Differentiate:

    1)\;F(y)\,=\,\left(\frac{1}{y^2}-\frac{3}{y^4}\right)(y+5y^3)

    2)\;y\,=\,\frac{t^2}{3t^2-2t+1}

    3)\;y\,=\,\frac{v^3-2v\sqrt{v}}{v}

    Simplify first . . .

    1)\;\;F(y)\;=\;(y^{-2} - 3y^{-4})(y + 5y^3)\;= \;y^{-1} + 5y - 3y^{-3} - 15y^{-1}\;= 5y - 3y^{-3} - 14y^{-1}


    3)\;\;y\;=\;\frac{v^3 - 2v\sqrt{v}}{v} \;= \;\frac{v^3 - 2v^{\frac{3}{2}}}{v} \;= v^2 - 2v^{\frac{1}{2}}


    #2 doesn't simplify, but you applied the Quotient Rule correctly . . . good work!


    [By the way, pay no attention to the man behind the curtain.
    Partial Fractions are used mostly for Integration, never for Differentiation.]
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  7. #7
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    I was confused with integration with #2. Sorry, I was just trying to help.

    I'm the idiot behind the curtain, never pay attention to me.

    KK
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