If $f(x)=\ln(\sqrt{x})$, then $f''(x)=$

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Rewrite $f(x)$ into fractional exponent form.

$$f(x)=\ln(\sqrt{x})=\ln{x^{\frac{1}{2}}}$$ $$f(x)= \frac{1}{2}\ln{x} \tag*{\tiny log of power rule}$$

Calculating the first derivative,

$$f'(x)= \frac{1}{2}\cdot \frac{1}{x}$$ $$f'(x)= \frac{1}{2x}$$

Again, rewrite into fractional exponent form to calculate the second derivative,

$$f''(x)=\frac{1}{2}x^{-1}$$ $$=-1 \cdot \frac{1}{2} \cdot x^{-2}$$ $$= \boxed{-\frac{1}{2x^2}}$$