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Math Help - Diff questions

  1. #1
    NeF
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    Question Diff questions

    I want to double check these answers, this book doesnt supply answers.

    1
     y= \frac{5x^4-3x}{x^2}\
     y= 4x^2-2x^-1
     \frac{dy}{dx}\ = 8x+2x^-2
     8x+\frac{2}{x^2}\

    2
     f(x)=(\frac{1}{x^2}+2)(x-3)
     f(x)=(x^-2+2)(x-3)
     f(x)= x^-1-3x^2+2x-6
     f'(x)= x^-1-6x^-3+2
     =  \frac{-1}{x^2}\ + \frac{6}{x^3}+2

    3
     y= \frac{2\sqrt(x)}{x}
     y= 2x^\frac{-1}{2}
     \frac{dy}{dx} = -x^\frac{-3}{2}
     = -\frac{1}{x^\frac{3}{2}}

    Thanx in advance
    Last edited by NeF; July 24th 2006 at 12:18 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by NeF
    I want to double check these answers, this book doesnt supply answers.

    1
     y= \frac{5x^4-3x}{x^2}\
     y= 4x^2-2x^-1
     \frac{dy}{dx}\ = 8x+2x^-2
     8x+\frac{2}{x^2}\
    Your second line is wrong. Given your second line, your third and fourth lines are correct.
    y = 5x^2 - 3x^{-1}

    Since you've evidently got the rules right, I'll let you finish it.

    -Dan
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by NeF
    I want to double check these answers, this book doesnt supply answers.

    1
     y= \frac{5x^4-3x}{x^2}\
     y= 4x^2-2x^-1
     \frac{dy}{dx}\ = 8x+2x^-2
     8x+\frac{2}{x^2}\
    It's not entirely clear what you are asking, but here I will assume that
    you want the derivative of:

     y= \frac{5x^4-3x}{x^2}\

    First divide through on the right hand side by the x^2 at the bottom:

     y= 5x^2-3x^{-1}\

    Now differentiate using the rules for differentiating powers:

     \frac{dy}{dx}\ = 10x+3x^{-2}

    Now rearrange and simplify (back towards something like the form we started
    from):

     \frac{dy}{dx}\ = 10x+\frac{3}{x^{2}}

     \frac{dy}{dx}\ = \frac{10x^3+3}{x^{2}}

    RonL
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by NeF
    I want to double check these answers, this book doesnt supply answers.

    2
     f(x)=(\frac{1}{x^2}+2)(x-3)
     f(x)=(x^-2+2)(x-3)
     f(x)= x^-1-3x^2+2x-6
     f'(x)= x^-1-6x^-3+2
     =  \frac{-1}{x^2}\ + \frac{6}{x^3}+2
    Your third line is wrong. (Typo?) Given that then your derivative is wrong. However, your final answer is correct.
     f(x)= x^{-1}-3x^{-2}+2x-6
     f'(x)= -x^{-2}+6x^{-3}+2

    Also, you might look into the product rule to do this one:
     f(x)=(\frac{1}{x^2}+2)(x-3)
     f'(x) = \frac{-2}{x^3}(x-3) + \left ( \frac{1}{x^2}+2 \right ) \cdot 1
     f'(x) = \frac{-2x + 3}{x^3} + \frac{1+2x^2}{x^2}
    etc, etc...

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by NeF
    I want to double check these answers, this book doesnt supply answers.

    3
     y= \frac{2\sqrt(x)}{x}
     y= 2x^\frac{-1}{2}
     \frac{dy}{dx} = -x^\frac{-3}{2}
     = -\frac{1}{x^\frac{3}{2}}
    This one looks okay. By the way, note the use of the {} in the Tex in the preceeding messages.

    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Oh! Hi there Captain!

    -Dan
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  7. #7
    NeF
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    Hectic, Thanks alot again for seeing all the stupid mistakes I am making.
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  8. #8
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    Quote Originally Posted by NeF
    Hectic, Thanks alot again for seeing all the stupid mistakes I am making.
    Does that mean you are going to kill yourself now!
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  9. #9
    NeF
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    No sadly I am not , I want to waste more of your precious air...

    That line refers to morons that do drugs and crap.
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  10. #10
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    Quote Originally Posted by NeF
    That line refers to morons that do drugs and crap.
    Then why do you have a picture of Raiden over there? You clan thing?
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  11. #11
    NeF
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    Yep thats for my clan
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