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    I would really appreciate it if someone could guide me through differentiating the following function (sqrtx+1)^2/x with the aid of the chain rule !THANK YOU SO MUCH YOU KIND PERSON!
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    (Original post by irenee)
    I would really appreciate it if someone could guide me through differentiating the following function (sqrtx+1)^2/x with the aid of the chain rule !THANK YOU SO MUCH YOU KIND PERSON!
     \displaystyle \frac{(\sqrt{x}+1)^2}{x} (please )

    or  \displaystyle (\sqrt{x}+1)^{\frac{2}{x}} (no please :afraid:)
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    (Original post by DylanJ42)
     \displaystyle \frac{(\sqrt{x}+1)^2}{x} (please )

    or  \displaystyle (\sqrt{x}+1)^{\frac{2}{x}} (no please :afraid:)
    The second could be done if you take logs of both sides and differentiate implicitly. First one is just normal quotient rule, although you could also use product rule.
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    (Original post by B_9710)
    The second could be done if you take logs of both sides and differentiate implicitly. First one is just normal quotient rule, although you could also use product rule.
    oh i know, but explaining it step by step and latexing it all would be not so fun :laugh:
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\dfrac{dy}{ydx} = \dfrac{1}{x \sqrt{x} (\sqrt{x} + 1)} - \dfrac{2ln(\sqrt{x} + 1)}{x^2}
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    (Original post by irenee)
    I would really appreciate it if someone could guide me through differentiating the following function (sqrtx+1)^2/x with the aid of the chain rule !THANK YOU SO MUCH YOU KIND PERSON!
    (Original post by DylanJ42)
     (1) \quad\displaystyle \frac{(\sqrt{x}+1)^2}{x} (please )

    or (2) \quad \displaystyle (\sqrt{x}+1)^{\frac{2}{x}} (no please :afraid:)
    if its (1), as im assuming it will be;


     \displaystyle \frac{d}{dx} \left(\frac{\sqrt{x} + 1)^2}{x} \right)

    Using product rule; (and call the above line  \displaystyle \frac{dy}{dx} )


     \displaystyle \frac{dy}{dx} = \frac{\left(x \cdot \frac{d}{dx}((\sqrt{x}+1)^2 \right) - \left((\sqrt{x} +1)^2 \cdot \frac{d}{dx}(x)\right)}{x^2}


    Now taking and working out  \displaystyle \frac{d}{dx} \left( (\sqrt{x}+1)^2 \right) first (as the rest is fairly standard);


    Let's assign a few variables first  \displaystyle \: z = (\sqrt{x}+1)^2,\quad u = \sqrt{x} +1, \quad \therefore z = u^2


     \displaystyle \text{Then}\: \frac{dz}{dx} = \frac{dz}{du} \cdot \frac{du}{dx}


     \displaystyle \frac{dz}{dx}  =  2u \cdot \frac{1}{2\sqrt{x}} = 2(\sqrt{x} +1) \cdot \frac{1}{2\sqrt{x}}  = 1 + \frac{1}{\sqrt{x}}


    So going back to;


     \displaystyle \frac{dy}{dx} = \frac{\left(x \cdot \frac{d}{dx}((\sqrt{x}+1)^2 \right) - \left((\sqrt{x} +1)^2 \cdot \frac{d}{dx}(x)\right)}{x^2}


    and subbing in the result we obtained above we get;


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


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


    the RHS of the top line multiples out and we get;

     \displaystyle \frac{\left(x + \sqrt{x} \right) - \left(x +2\sqrt{x} + 1 \right)}{x^2}


    which finally leaves us with;

     \displaystyle \frac{dy}{dx} = \frac{-1 - \sqrt{x}}{x^2}

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    in hindsight i wish it was (2) :laugh:
 
 
 

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