Answer
$f^{(n)}(x)=\frac{(-1)^{n+1}(n-1)!}{x^{n}}$
Work Step by Step
$$f^{(1)}(x)=(\ln(2x))'=\frac{(2x)'}{2x}=\frac{1}{x}=\frac{(1-1)!}{x}$$
$$f^{(2)}(x)=\left(\frac{1}{x}\right)'=\frac{(1)'x-1\cdot (x)'}{(x)^{2}}=\frac{-1}{x^{2}}=\frac{-(2-1)!}{x^{2}}$$
$$f^{(3)}(x)=\left(-\frac{1}{x^{2}}\right)'=-\frac{(1)'x^{2}-1\cdot (x^{2})'}{(x^{2})^{2}}=-\frac{-2x}{x^{4}}=\frac{2}{x^{3}}=\frac{(3-1)!}{x^{3}}$$
Using the same process the fourth derivative of $f$ is:
$$f^{(4)}(x)=\frac{-(4-1)!}{x^{4}}$$
Following the pattern it follows that:
$$f^{(n)}(x)=\frac{(-1)^{n+1}(n-1)!}{x^{n}}$$
