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Homework Help: Find derivative of Square root (x + square root(x + x^(1/2))) Help!

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Define f(x)=[itex]\sqrt{}(x + (\sqrt{}(x + \sqrt{}x)[/itex]
    Determine where f is differentiable and compute the derivative

    2. Relevant equations

    f'(xo)= lim as x approaches xo (f(x) - f(xo))/(x - xo)

    3. The attempt at a solution
    By the definition, f(x) = [itex]\sqrt{}x[/itex] does not have a derivative at 0, since limit as x approaches zero, x>0, is 1/[itex]\sqrt{}x[/itex], so can't have zero in denominator, not differentiable at 0.

    I think above function, because of Algebra of Derivatives would also not be differentiable at zero. I keep ending up with a big mess trying to find it's derivative, I think I'm missing a rule somewhere, any suggestions are welcome.
  2. jcsd
  3. Mar 27, 2012 #2
    Chain rule can be thought about as fractions:
    [tex] f(x) = \sqrt{x+\sqrt{x+\sqrt{x}}}[/tex]
    and we want
    [tex] \frac{df}{dx} = f'(x)[/tex]
    then let
    [tex] g(x) = x + \sqrt{x+\sqrt{x}}[/tex]


    [tex] f(g(x)) = \sqrt{g(x)}[/tex]

    [tex] \frac{df}{dg}\frac{dg}{dx} = \frac{df}{dx} = g'(x)\frac{1}{2}g(x)^{-\frac{1}{2}}[/tex]

    Let me start you off on the next step and see if you can finish it. Let
    [tex] h(x) = \sqrt{x+\sqrt{x}}[/tex]
    [tex] g(x) = x + h(x)[/tex]
    [tex] \frac{dg}{dx} = 1 + \frac{dh}{dx}[/tex]

    Hint: use the same logic already presented to find the derivative of h(x). Then substitute all of your answers into previous equations for the final answer.
  4. Mar 27, 2012 #3
    I think the limit as x -> xo, if you multiply numerator and denominator by it conjugate you get, f'(xo) = 1/2(xo+(xo + (xo)^(1/2))^(1/2))^(1/2))

    Is that it?
  5. Mar 27, 2012 #4
    I apologize. I didn't see you were dealing with the limit.

    [tex] f(x) = \sqrt{x+\sqrt{x+\sqrt{x}}}[/tex]

    Then we seek

    [tex] Z = \frac{f(x)-f(x_0)}{x-x_0}=\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x_0+\sqrt{x_0+\sqrt{x_0}}}}{x-x_0}[/tex]

    but let
    [tex] g(x) = x + \sqrt{x+\sqrt{x}}[/tex]


    [tex]Z = \frac{\sqrt{g(x)} - \sqrt{g(x_0)}}{g(x)-g(x_0)}\frac{g(x)-g(x_0)}{x-x_0}[/tex]

    But let
    [tex] h(x) = \sqrt{x+\sqrt{x}}[/tex]


    [tex]Z = \frac{\sqrt{g(x)} - \sqrt{g(x_0)}}{g(x)-g(x_0)}\frac{x+h(x)-(x_0+h(x_0))}{x-x_0}=\frac{\sqrt{g(x)} - \sqrt{g(x_0)}}{g(x)-g(x_0)} \left ( \frac{x-x_0}{x-x_0} + \frac{h(x) - h(x_0)}{x-x_0} \right )[/tex]

    But let

    [tex] y(x) = x + \sqrt{x}[/tex]


    [tex]Z =\frac{\sqrt{g(x)} - \sqrt{g(x_0)}}{g(x)-g(x_0)} \left ( \frac{x-x_0}{x-x_0} + \frac{\sqrt{y(x)} - \sqrt{y(x_0)}}{y(x)-y(x_0)}\frac{y(x)-y(x_0)}{x-x_0} \right )[/tex]
    and the final one falls into place:

    [tex]Z =\frac{\sqrt{g(x)} - \sqrt{g(x_0)}}{g(x)-g(x_0)} \left ( \frac{x-x_0}{x-x_0} + \frac{\sqrt{y(x)} - \sqrt{y(x_0)}}{y(x)-y(x_0)}\left ( \frac{x-x_0}{x-x_0} +\frac{\sqrt{x}-\sqrt{x_0}}{x-x_0} \right ) \right )[/tex]

    Taken the limit of Z as x approaches x_0. Do so for each piece.
  6. Mar 28, 2012 #5
    If you don't know how to handle the squareroot case frequent in that final equation I wrote, consider x - x_0 as a difference of squares.
  7. Mar 28, 2012 #6
    Thank you. Your approach makes sense and is clear.
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