Answer
$f'(x)=\frac{-1}{2x\sqrt{1-\ln(x)}}$
$D:x \in \mathbb R: 0 \lt x \leq e $
Work Step by Step
Using the chain rule it follows:
$$f'(x)=\frac{(1-\ln(x))'}{2\sqrt{1-\ln(x)}}$$
$$f'(x)=\frac{0-\frac{1}{x}}{2\sqrt{1-\ln(x)}}$$
$$f'(x)=\frac{-1}{2x\sqrt{1-\ln(x)}}$$
The domain of $f$ is:
$$D:x \in \mathbb R: 1-\ln(x) \geq 0~~\text{and}~~x \gt 0$$
$$D:x \in \mathbb R: -\ln(x) \geq -1~~\text{and}~~x \gt 0$$
$$D:x \in \mathbb R: \ln(x) \leq 1~~\text{and}~~x \gt 0$$
$$D:x \in \mathbb R: x \leq e ~~\text{and}~~x \gt 0$$
$$D:x \in \mathbb R: 0 \lt x \leq e $$
