If \(x=\log _e\left(\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}\right), \tan \frac{y}{2}=\sqrt{\frac{1-t}{1+t}}\). Then \(\left(y_1\right)_{t=\frac{1}{2}}\) has the value

\(\begin{aligned} & x=\log _{\mathrm{e}}\left(\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}\right) \\ & \Rightarrow \mathrm{e}^x=\frac{1-\tan \frac{y}{2}}{1+\tan \frac{y}{2}}\end{aligned}\)
\(\Rightarrow \mathrm{e}^x=\tan \left(\frac{\pi}{4}-\frac{y}{2}\right) \quad\)....(i) \(\left[\because \tan \frac{\pi}{4}=1\right]\)
Differentiating w.r.t. \(x\), we get
\(\mathrm{e}^x=\sec ^2\left(\frac{\pi}{4}-\frac{y}{2}\right) \cdot\left(\frac{-1}{2}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}\)
\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-2 \mathrm{e}^x \cos ^2\left(\frac{\pi}{4}-\frac{y}{2}\right)\)
\(\begin{aligned} & \text { When } \mathrm{t}=\frac{1}{2}, \\ & \tan \frac{y}{2}=\sqrt{\frac{1-\frac{1}{2}}{1+\frac{1}{2}}} \\ & \Rightarrow \tan \frac{y}{2}=\frac{1}{\sqrt{3}} \\ & \Rightarrow \frac{y}{2}=\frac{\pi}{6}\end{aligned}\)
Substituting \(\frac{y}{2}=\frac{\pi}{6}\) in (i), we get
\(\mathrm{e}^x=\tan \frac{\pi}{12}=2-\sqrt{3}\)
\(\therefore \quad\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{\mathrm{t}=\frac{1}{2}}=-2(2-\sqrt{3}) \cos ^2 \frac{\pi}{12}\)
\(\begin{aligned} & =-2(2-\sqrt{3})\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)^2 \\ & =\frac{-1}{2}(2-\sqrt{3})(2+\sqrt{3}) \\ & =-\frac{1}{2}\end{aligned}\)