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Math Help - y= sqrt2x + 2sqrt2x

  1. #1
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    Angry y= sqrt2x + 2sqrt2x

    y= sqrt2x + 2sqrt2x

    I need to find dy/dx,

    When i differentiate each part I get the same result which I know is not correct,

    y = 2x^-1/2 + 2(x^-1/2)

    dy/dx = 1/2sqrtx + 1/2sqrtx

    any help would be good
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  2. #2
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    Smile

    Quote Originally Posted by drb11 View Post
    y= sqrt2x + 2sqrt2x

    I need to find dy/dx,

    When i differentiate each part I get the same result which I know is not correct,

    y = 2x^-1/2 + 2(x^-1/2) \longleftarrow should be y = (2x)^{1/2} + 2(2x)^{1/2}
    . . .
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  3. #3
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    well using the power rule:

    f'(x)=nx^{n-1}

    We see in your problem:

    y= \sqrt{2x} + 2\sqrt{2x}
    1.
    y= (2x)^{1/2}+2(2x)^{1/2}
    2.
    <br />
y= (x)^{-1/2}+2(x)^{-1/2}
    3.
    y= x^{-1/2}+2x^{-1/2}
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  4. #4
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    [quote=Zanderist;475964]well using the power rule:

    f'(x)=nx^{n-1}

    We see in your problem:

    y= \sqrt{2x} + 2\sqrt{2x}

    y= (2x)^{1/2}+2(2x)^{1/2}

    <br />
y'= (x)^{-1/2}+2(x)^{-1/2} \longleftarrow should be y'= (2x)^{-1/2}+2(2x)^{-1/2}

    [\quote]

    . . .
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  5. #5
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    You are aware that \sqrt{2x}+ 2\sqrt{2x}= 3\sqrt{2x} aren't you? Its derivative is \frac{3}{2}\sqrt{2}x^{-1/2}= \frac{3}{2}\sqrt{\frac{2}{x}}.
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  6. #6
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    oops, it seems I made a mistake when posting the initial problem,

    it was supposed to be:

    y = sqrt2x + 2sqrtx

    sorry about that

    I know what the answer is but I don't know how to get there....
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by drb11 View Post
    oops, it seems I made a mistake when posting the initial problem,

    it was supposed to be:

    y = sqrt2x + 2sqrtx

    sorry about that

    I know what the answer is but I don't know how to get there....
    \sqrt{2x}+2\sqrt{x}=(\sqrt{2}+2)\sqrt{x}
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  8. #8
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    Sweet, thanks
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by drb11 View Post

    I know what the answer is but I don't know how to get there....
    We see people writing this all the time. However you do not know what the answer is all you know is what it says at the back of the book. Which is not the same thing at all.

    CB
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  10. #10
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    Sorry captainblack sparrow. The reason I asked the question; for help on the process on how to get to the correct solution.I thought this was a math help forum ,obviously not, oh well, please revoke my membership. Feel free to feed upon my ego
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  11. #11
    Grand Panjandrum
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    Quote Originally Posted by drb11 View Post
    Sorry captainblack sparrow. The reason I asked the question; for help on the process on how to get to the correct solution.I thought this was a math help forum ,obviously not, oh well, please revoke my membership. Feel free to feed upon my ego
    You had your help.

    But it seems that if we comment on your sloppy use of language you take your balls back and run off. Now who is the loser, you want help, we give it (or rather DeDust, Halls, Zanderist and Drexel28 give it), but you won't play any more because I wish to correct a ridiculous use of language that I see all the time.

    Ohhh.. well, my loss I suppose. Goodbye

    CB
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