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    Differentiate  \frac{x}{\sqrt{x+3}} simplifying your answer as far as possible.

    I've gotten a little stuck.
    Here is the solution:
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    \frac{x+6}{2(x+3)^\frac{3}{2}}


    my working is as follows:
    let  u = x and  v = \sqrt{x+3}

    \frac{du}{dx} = 1 and  \frac{dv}{dx} = \frac{1}{2}(x+3)^\frac{-1}{2}

    thus using quotient rule:  \frac{\frac{x}{2\sqrt{x+3}} - \sqrt{x+3}}{\sqrt{x+3}^2} = \frac{\frac{x}{2\sqrt{x+3}} - \sqrt{x+3}}{x+3}

    Where do I go from here?
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    (Original post by purplefrog)
    Differentiate  \frac{x}{\sqrt{x+3}} simplifying your answer as far as possible.

    I've gotten a little stuck.
    Here is the solution:
    Spoiler:
    Show
    \frac{x+6}{2(x+3)^\frac{3}{2}}


    my working is as follows:
    let  u = x and  v = \sqrt{x+3}

    \frac{du}{dx} = 1 and  \frac{dv}{dx} = \frac{1}{2}(x+3)^\frac{-1}{2}

    thus using quotient rule:  \frac{\frac{x}{2\sqrt{x+3}} - \sqrt{x+3}}{\sqrt{x+3}^2} = \frac{\frac{x}{2\sqrt{x+3}} - \sqrt{x+3}}{x+3}

    Where do I go from here?
    You made a small mistake with which way round the quotient rule goes. You should have got  \frac{\sqrt{x+3} -\frac{x}{2\sqrt{x+3}} }{\sqrt{x+3}^2}. Do that same simplification step you did before, then multiply top and bottom of the fraction by \sqrt{x+3}.
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    your numerator is the wrongway round .... Wherever v is it goest before the U..
    can't really explain it but it should be Root(x+3)-(x/2root(x+3) and then divide by v^2

    ^^ ahh he explained it clearer then i ever could have
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    If you hate the quotient rule more than the product rule, then you could always just use the product rule. A question will never specifically ask you to use one method over the other.

    For me, turning  \frac{x}{\sqrt{x+3}} into x(x+3)^{-\frac{1}{2}} and using the product (and chain) rule is simpler and faster than trying to remember the quotient rule.

    Just an idea though. Use whatever you feel more comfortable with.
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    (Original post by purplefrog)
    x
    I'm going to start from scratch so it makes sense to me

     \int \frac {x}{\sqrt{x+3}}dx = \int \frac {u}{v} = \frac {v \cdot \frac {du}{dx} - u \cdot \frac {dv}{dx}}{v^2}

     u = x \ \ \ \ v = \sqrt{x+3} = {(x+3)}^{\frac {1}{2}}

     \frac {du}{dx} = 1 \ \ \ \ \ \frac {dv}{dx} = \frac {{(x+3)}^{- \frac {1}{2}}}{2} = \frac {1}{2 \sqrt{x+3}}

     \int \frac {x}{\sqrt{x+3}}dx =\frac {\sqrt{x+3} -  \frac {x}{2 \sqrt {x+3}}}{{\sqrt {x+3}}^2}

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

     \frac {x+6}{2(x+3)^{\frac {3}{2}}}
 
 
 
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