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    • Thread Starter
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    Use the chain rule to differentiate:
    y= \left( \sqrt x + \frac{1}{\sqrt x} \right)^3.

    I think i've managed to work through it but some confirmation of whether I'm correct or not would be nice. I got:
    \frac{dy}{dx} = \left( \frac{3}{2\sqrt x} - \frac{3}{2 (\sqrt x)^3} \right) \left( \sqrt x + \frac{1}{ \sqrt x} \right)^2

    Do you think that this question needs to be put into a different form or if it can be simplified any further?
    Thanks in advance.
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    Well you could factorise \dfrac{3}{2} out of the first bracket, but other than that I don't see why it should be simplified any further. If you wanted to be really silly about it you could take \dfrac{1}{x} out as well, meaning you get:
    \dfrac{3}{2x} \left( \sqrt{x} - \dfrac{1}{\sqrt{x}} \right) \left( \sqrt{x} + \dfrac{1}{\sqrt{x}} \right)^2
     = \dfrac{3}{2x}\left( x - \dfrac{1}{x} \right) \left( \sqrt{x} + \dfrac{1}{\sqrt{x}} \right)
    But that's pushing it.
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    (Original post by nuodai)
    Well you could factorise \dfrac{3}{2} out of the first bracket, but other than that I don't see why it should be simplified any further. If you wanted to be really silly about it you could take \dfrac{1}{x} out as well, meaning you get:
    \dfrac{3}{2x} \left( \sqrt{x} - \dfrac{1}{\sqrt{x}} \right) \left( \sqrt{x} + \dfrac{1}{\sqrt{x}} \right)^2
     = \dfrac{3}{2x}\left( x - \dfrac{1}{x} \right) \left( \sqrt{x} + \dfrac{1}{\sqrt{x}} \right)
    But that's pushing it.
    Seconded.
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    (Original post by Farhan.Hanif93)
    Use the chain rule to differentiate:
    y= \left( \sqrt x + \frac{1}{\sqrt x} \right)^3.

    I think i've managed to work through it but some confirmation of whether I'm correct or not would be nice. I got:
    \frac{dy}{dx} = \left( \frac{3}{2\sqrt x} - \frac{3}{2 (\sqrt x)^3} \right) \left( \sqrt x + \frac{1}{ \sqrt x} \right)^2

    Do you think that this question needs to be put into a different form or if it can be simplified any further?
    Thanks in advance.
    Looks good to me.

    Am I dreaming or can it go to
    \frac{dy}{dx} = \frac{3}{2}  ( 1 - \frac{1}{  x^2} )
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    No - I'm dreaming - I missed the square on the final bracket. Sorry
    • Thread Starter
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    Thanks for the help guys
 
 
 
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