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    Given that y = 2x root(x) + 2 root(x) / x


    Show that dy/dx = x-1 / x root(x)

    Any help?
    Thanks
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    simplify the terms by combining the powers ? remember that the square root power is 1/2 ?
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    (Original post by orange8120)
    Given that y = 2x root(x) + 2 root(x)
    ----------------------------
    x

    Show that dy/dx = x-1
    ------------
    x root(x)

    Any help?
    Thanks
    Simplify y to: \displaystyle y = \frac{2x\sqrt{x} + 2\sqrt{x}}{x} = \frac{2x \sqrt{2}}{x} + \frac{2\sqrt{x}}{x} = 2\sqrt{x} + \frac{2}{\sqrt{x}} = 2x^{1/2} + 2x^{-1/2} then differentiate.
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    (Original post by Zacken)
    Simplify y to: \displaystyle y = \frac{2x\sqrt{x} + 2\sqrt{x}}{x} = \frac{2x \sqrt{2}}{x} + \frac{2\sqrt{x}}{x} = 2\sqrt{x} + \frac{2}{\sqrt{x}} = 2x^{1/2} + 2x^{-1/2} then differentiate.
    So after I differentiate I get x^-1/2 - x ^-3/2.
    How do I simplify that to x-1 / x root(x)

    Many thanks
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    (Original post by orange8120)
    So after I differentiate I get x^-1/2 - x ^-3/2.
    How do I simplify that to x-1 / x root(x)

    Many thanks
    So you have: \displaystyle \frac{1}{\sqrt{x}} - \frac{1}{x\sqrt{x}} = \frac{x}{x\sqrt{x}} - \frac{1}{x\sqrt{x}} = \cdots

    Remember that: x^{3/2} = x \cdot x^{1/2} = x \sqrt{x}
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    (Original post by Zacken)
    So you have: \displaystyle \frac{1}{\sqrt{x}} - \frac{1}{x\sqrt{x}} = \frac{x}{x\sqrt{x}} - \frac{1}{x\sqrt{x}} = \cdots

    Remember that: x^{3/2} = x \cdot x^{1/2} = x \sqrt{x}
    Thanks a lot. Also i'm strugglng with:

    root(x) - 1 / x^2

    Find gradient when x = 2.
    I get that i need to differentiate and then sub in 2 as x but no matter how many times I do it I can't get the right answer.
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    (Original post by orange8120)
    Thanks a lot. Also i'm strugglng with:

    root(x) - 1 / x^2

    Find gradient when x = 2.
    I get that i need to differentiate and then sub in 2 as x but no matter how many times I do it I can't get the right answer.
    Again, first thing first - simplify, then differentiate:

    \displaystyle \frac{\sqrt{x} - 1}{x^2} = \frac{\sqrt{x}}{x^2} - \frac{1}{x^2} = x^{-3/2} - x^{-2}

    What do you get when you differentiate that?
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    (Original post by Zacken)
    Again, first thing first - simplify, then differentiate:

    \displaystyle \frac{\sqrt{x} - 1}{x^2} = \frac{\sqrt{x}}{x^2} - \frac{1}{x^2} = x^{-3/2} - x^{-2}

    What do you get when you differentiate that?
    -3/2x +2x^-3
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    (Original post by orange8120)
    -3/2x +2x^-3
    See bolded bit. You need to subtract one from the power. What is \displaystyle x^{-3/2 - 1}?

    What's \displaystyle -\frac{3}{2} - 1 = -1.5 - 1?
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    (Original post by Zacken)
    See bolded bit. You need to subtract one from the power. What is \displaystyle x^{-3/2 - 1}?

    What's \displaystyle -\frac{3}{2} - 1 = -1.5 - 1?
    -3/2x^-5/2 + 2x^-3
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    (Original post by orange8120)
    -3/2x^-5/2 + 2x^-3
    And what do you get when you put x=2 into that?
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    (Original post by Zacken)
    And what do you get when you put x=2 into that?
    -3/2(2^-5/2) + 2(2^-3) = -0.0152
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    (Original post by orange8120)
    -3/2(2^-5/2) + 2(2^-3) = -0.0152
    That's correct, in't it?
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    (Original post by Zacken)
    That's correct, in't it?
    Yes it is! Thank you so so much!
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    (Original post by orange8120)
    Yes it is! Thank you so so much!
    No problemo!
 
 
 
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