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  1. #1
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    derivate

    please try to do these questions.
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  2. #2
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    Quote Originally Posted by m777 View Post
    please try to do these questions.
    1. Find the derivative of (2 \sqrt{x} + 2)(x - \sqrt{x}).

    This uses the product rule, so:
    \frac{d}{dx}[(2 \sqrt{x} + 2)(x - \sqrt{x})]

    = \left ( 2 \cdot \frac{1}{2} \frac{1}{\sqrt{x}} \right )(x - \sqrt{x}) + (2 \sqrt{x} + 2) \left ( 1 - \frac{1}{2} \frac{1}{\sqrt{x}} \right )

    = \left ( \frac{1}{\sqrt{x}} \right )(x - \sqrt{x}) + (\sqrt{x} + 1) \left ( 2 - \frac{1}{\sqrt{x}} \right ) (I've just moved around the 2 in the second term here.)

    = \frac{x}{\sqrt{x}} - 1 + 2\sqrt{x} - 1 + 2 - \frac{1}{\sqrt{x}}

    = 3\sqrt{x} - \frac{1}{\sqrt{x}}

    = \frac{3x - 1}{\sqrt{x}}

    -Dan
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  3. #3
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    nr. 3 only

    Quote Originally Posted by m777 View Post
    please try to do these questions.
    Hello, m777,

    f(x)=\left( \sqrt{x}-\frac{1}{\sqrt{x}} \right)^2. Thus using chain rule

    f'(x)=2 \cdot \left( \sqrt{x}-\frac{1}{\sqrt{x}} \right) \cdot \left( \frac{1}{2 \sqrt{x}}+\frac{1}{2} \cdot x^{-\frac{3}{2}} \right)

    f'(x)=2 \cdot \left( \sqrt{x}-\frac{1}{\sqrt{x}} \right) \cdot \frac{1}{2} \left( \frac{1}{\sqrt{x}}+ \frac{1}{(\sqrt{x})^3} \right)

    Expand the paranthese:

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

    EB
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  4. #4
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    Hello, m777!

    1) Prove that the derivative of . y \:=\:\left(2\sqrt{x} + 2\right)\left(x - \sqrt{x}\right) .is: . \frac{dy}{dx} \:=\:\frac{3x - 1}{\sqrt{x}}

    Multiply first: . y \:=\:\left(2\sqrt{x} + 2\right)\left(x - \sqrt{x}\right) \;=\;2x\sqrt{x} - 2x + 2x - 2\sqrt{x}<br />

    And we have: . y \;= \;2x^{\frac{3}{2}} - 2x^{\frac{1}{2}}

    Then: . \frac{dy}{dx}\;=\;2\cdot\frac{3}{2}x^{\frac{1}{2}} - 2\cdot\frac{1}{2}x^{-\frac{1}{2}} \;= \;3x^{\frac{3}{2}} - x^{-\frac{1}{2}}

    Multiply "top and bottom" by x^{\frac{1}{2}}:\;\;\frac{dy}{dx}\;= \;\frac{x^{\frac{1}{2}}}{x^{\frac{1}{2}}}\cdot\fra  c{3x^{\frac{1}{2}} - x^{-\frac{1}{2}}}{1} \;= \;\frac{3x - 1}{x^{\frac{1}{2}}}

    Therefore: . \boxed{\frac{dy}{dx} \;= \;\frac{3x - 1}{\sqrt{x}}}



    3) Differentiate: . y \:=\:\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2

    Square first: . y \;= \;\left(\sqrt{x}\right)^2 - 2\left(\sqrt{x}\right)\left(\frac{1}{\sqrt{x}}\rig  ht) + \left(\frac{1}{\sqrt{x}}\right)^2 \;= \;x - 2 + \frac{1}{x}

    So we have: . y \;= \;x - 2 + x^{-1}

    Then: . \frac{dy}{dx} \;= \;1 - x^{-2} \;= \;1 - \frac{1}{x^2}

    Therefore: . \boxed{\frac{dy}{dx} \;= \;\frac{x^2 - 1}{x^2}}



    2) Differentiate: . y \;= \;\frac{\sqrt{a+x} + \sqrt{a - x}}{\sqrt{a+x} - \sqrt{a-x}}

    I'm still working on this one, but I would "rationalize" first.


    Multiply top and bottom by \left(\sqrt{a+x} + \sqrt{a-x}\right)

    . . y \;= \;\frac{\sqrt{a+x} + \sqrt{a - x}}{\sqrt{a+x} + \sqrt{a - x}}\cdot\frac{\sqrt{a+x} + \sqrt{a+x}}{\sqrt{a-x} - \sqrt{a-x}} \;=\;\frac{\left(\sqrt{a+x} + \sqrt{a-x}\right)^2}{\left(\sqrt{a+x}\right)^2 - \left(\sqrt{a-x}\right)^2}

    . . y \;= \;\frac{a + x + 2\sqrt{a+x}\sqrt{a-x} + a - x}{(a+x) - (a - x)} \;=\;\frac{2a + 2\sqrt{a^2-x^2}}{2x}

    . . y \;= \;\frac{a + \sqrt{a^2-x^2}}{x} . . . then differentiate.

    . . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    I don't agree with the given answer.
    . . (Why aren't the 2's cancelled out?)

    Could it be: . \frac{dy}{dx}\;= \;\frac{a}{a - \sqrt{a^2-x^2}} ?

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  5. #5
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    answer

    hello,
    soroban
    In my book answer is same as i mention in question.any how i will concern with my teacher but i agree with your answer.
    many best wishes for you
    from
    m777.
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