Question
${d \over {dx}}{\log _{\sqrt x }}(1/x)$ is equal to
$= \frac{{\log \left( {\frac{1}{x}} \right)}}{{\log \sqrt x }} $
$= \frac{{( - 1)\log x}}{{(1/2)\,\log x}} = - 2$
==> $f'(x) = 0$.
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$8 \sqrt{x}(\sqrt{9+\sqrt{x}}) d y=(\sqrt{4+\sqrt{9+\sqrt{x}}})^{-1} d x, \quad x>0$
and $y(0)=\sqrt{7}$, then $y(256)=$