$\begin{aligned} & y=\log \sqrt{\tan x} \\ & \therefore \quad \frac{d y}{d x}=\frac{1}{\sqrt{\tan x}} \times \frac{1}{2 \sqrt{\tan x}} \times \sec ^2 x=\frac{\sec ^2 x}{2 \tan x} \\ & \therefore \quad\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{4}}=\frac{(\sqrt{2})^2}{2}=1\end{aligned}$