$\log \left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right)$
$\log \left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right)$
$=\log \left(\frac{1-\sin x}{\cos x}\right)$
$\begin{aligned} & =\log \left(\frac{1}{\cos x}-\frac{\sin x}{\cos x}\right) \\ & =\log (\sec x-\tan x)\end{aligned}$
Differentiating w.r.t. $x$, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}[\log (\sec x-\tan x)] \\ & =\frac{1}{\sec x-\tan x} \cdot \frac{d}{d x}(\sec x-\tan x) \\ & =\frac{1}{\sec x-\tan x} \times\left(\sec x \tan x-\sec ^2 x\right) \\ & =\frac{-\sec x(\sec x-\tan x)}{\sec x-\tan x} \\ & =-\sec x \\ & \end{aligned}$
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