$f(x) = \sqrt{x + \sqrt{x}} = (x + \sqrt{x})^{1/2}$

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Derivare la funzione

$f(x) = sqrt{x + sqrt{x}} = (x + sqrt{x})^{1/2}$ $f'(x) = frac{1}{2}cdot (x + sqrt{x})^{-1/2}cdot (1 + frac{1}{2sqrt{x}}) =$ $frac{1}{2 sqrt{x + sqrt{x}}}cdot frac{2sqrt{x} + 1}{2sqrt{x}} = frac{2x + sqrt{x}}{4sqrt{x + sqrt{x}}cdot x}$