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Compute the derivative of the given function

by RoRi

Compute the derivative of the function

    \[ \sqrt{x + \sqrt{x + \sqrt{x}}}. \]

To help our computation, let’s define g(x) = \sqrt{x + \sqrt{x}}. Then,

    \begin{align*} g'(x) &= \left( \frac{1}{2 \sqrt{x + \sqrt{x}}} \right) \left( 1 + \frac{1}{2 \sqrt{x}} \right) \\ &= \frac{1}{2 \sqrt{x + \sqrt{x}}} + \frac{1}{4 \sqrt{x} \sqrt{x + \sqrt{x}}} \\ & = \frac{1}{ 2 g(x)} + \frac{1}{(4 \sqrt{x}) g(x)}. \end{align*}

Then, using this g(x) and the chain rule, we have

    \begin{align*} f(x) &= \sqrt{x + g(x)} \\ \implies f'(x) &= \frac{1}{2 \sqrt{x + g(x)}} (1 + g'(x)) \\[9pt] &= \left( \frac{1}{2 \sqrt{x + g(x)}} \right) \left( 1 + \frac{1}{2 g(x)} + \frac{1}{ (4 \sqrt{x})g(x)} \right) \\[9pt] &= \frac{(4 \sqrt{x}) g(x) + 2 \sqrt{x} + 1}{(8 \sqrt{x}) g(x) \sqrt{x + g(x)}}. \end{align*}

(This is the form the answer in the back of the book gives, so we’ll leave it in this form. If you want, you can plug back in g(x) = \sqrt{x + \sqrt{x}}, but I don’t think it simplifies to anything nice.)

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