Answer
$f'(x) = \frac{2\ln x+1}{2\sqrt {\ln x}}$
Work Step by Step
$f(x) = x \sqrt{ \ln x} $
$f(x) = x(\ln x)^{\frac{1}{2}}$
$f'(x) = (1)(\ln x)^{\frac{1}{2}} + x(\frac{1}{2})(\ln x)^{-\frac{1}{2}}(\frac{1}{x})$
$f'(x) = (\ln x)^{\frac{1}{2}} + (\frac{x}{2x})(\ln x)^{-\frac{1}{2}}$
$f'(x) = (\ln x)^{\frac{1}{2}} + (\frac{1}{2})(\ln x)^{-\frac{1}{2}}$
$f'(x) = \sqrt {\ln x} + (\frac{1}{2})(\frac{1}{\sqrt {\ln x}})$
$f'(x) = \sqrt {\ln x} + (\frac{1}{2\sqrt {\ln x}})$
$f'(x) = \frac{(\sqrt {\ln x})({2\sqrt {\ln x}})+1}{2\sqrt {\ln x}}$
$f'(x) = \frac{2\ln x+1}{2\sqrt {\ln x}}$
