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Math Help - Chain Rule with Three Nested Square Roots

  1. #1
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    Chain Rule with Three Nested Square Roots

    This one makes my brain spin:

    Find the derivative:

    y = \sqrt{x + \sqrt{x + \sqrt{x}}}

    I see four composed functions here (sorry I'm running out of letters at the end):
    y = f(u) = \sqrt{u}
    u = g(x) = x + \sqrt{x}
    x = h(t) = t + \sqrt{t}
    t = z(v) = \sqrt {v}

    Since I probably have already made a mistake, I'll just stop here for now to see what people think of this.

    Thanks.
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  2. #2
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    Assuming that I have the functions identified correctly, the derivative calculation should work like this. This is really messy, so I'll split the derivatives out for now.

    y' = [f'(u)][g'(x)][h'(t)][z'(v)]

    f'(u) = [\frac{1}{2}(\sqrt{x+\sqrt{x+\sqrt{x}}}^{-1/2}]
    g'(x) = [1 + \frac{1}{2}x^{-1/2}]
    h'(t) = [1 + \frac{1}{2}t^{-1/2}]
    z'(v) = \frac{1}{2}v^{-1/2}

    Am I doing this right?
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  3. #3
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    \displaystyle y = \left[x + \left(x + x^{1/2})\right^{1/2}\right]^{1/2}.


    Let \displaystyle u = x + \left(x + x^{1/2}\right)^{1/2} so that \displaystyle y = u^{1/2}.

    \displaystyle \frac{dy}{du} = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{x + \sqrt{x + \sqrt{x}}}}.


    \displaystyle \frac{du}{dx} = 1 + \frac{d}{dx}\left[(x + x^{1/2})^{1/2}\right].

    Another application of the Chain Rule will lead you to \displaystyle \frac{du}{dx}, which you can then use to find \displaystyle \frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}.
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  4. #4
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    Thanks. I thought that I had this, but I still don't get the right answer (or so says the website). Here is what I wound up with.

    y' = [\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}][1 + \frac{1}{2\sqrt{x+\sqrt{x}}}][1+\frac{1}{2\sqrt{x}}]

    I must be still missing something, but at this point, I give up.

    Thanks for you help, though. I really appreciate it.
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    If f(x) = \sqrt{x+\sqrt{x+\sqrt{x}}}, then using [\sqrt{f(x)}]' = \frac{f'(x)}{2\sqrt{f(x)}} does it in three steps:

    \displaystyle f'(x) = \frac{\left(x+\sqrt{x+\sqrt{x}}\right)'}{2\sqrt{x+  \sqrt{x+\sqrt{x}}}} = \frac{1+\frac{\left(x+\sqrt{x}\right)'}{2\sqrt{x+\  sqrt{x}}}}{2\sqrt{x+\sqrt{x+\sqrt{x}}}} =  \frac{1+\frac{1+\frac{1}{2\sqrt{x}}}{2\sqrt{x+\sqr  t{x}}}}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}.

    It seems you only had a bracket mistake there though (a common issue with the chain rule).
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  6. #6
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    Quote Originally Posted by joatmon View Post
    y' = [\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}][1 + \frac{1}{2\sqrt{x+\sqrt{x}}}][1+\frac{1}{2\sqrt{x}}]
    Thanks for you help, though. I really appreciate it.
    I thought I should elaborate where your bracket mistake is. It should be:

    \displaystyle y' = \left[\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}\right]\left[1 + \frac{1}{2\sqrt{x+\sqrt{x}}}\left(1+\frac{1}{2\sqr  t{x}}\right)\right].
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