$x=(\log x)^{\tan y}$, then $\frac{d y}{d x}$ is

$\log x-\tan y$ $\log x+$ tan y $\frac{\log x-\tan y}{x \log x \sec ^2 y}$ $\frac{\log x-\tan y}{x \log x \sec ^2 y . \log \log x}$

$\frac{\log x-\tan y}{x \log x \sec ^2 y . \log \log x}$

$x=(\log x)^{\tan y} \Rightarrow \log x=\tan y \log \log x$ $\tan y=\frac{\log x}{\log \log x}$ Hence (4) is correct answer.