If ${\mathrm{e}}^{\mathrm{x}}=\frac{\sqrt{1+\mathrm{t}}-\sqrt{1-\mathrm{t}}}{\sqrt{1+\mathrm{t}}+\sqrt{1-\mathrm{t}}}$ and $\tan \mathrm{y}/2=\sqrt{\frac{1-\mathrm{t}}{1+\mathrm{t}}}$ then $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\mathrm{t}=1/2$ is

$$\begin{gathered} \begin{array}{rr}{e}^x=\frac{1-\tan \frac{y}{2}}{1+\tan \frac{y}{2}}=\tan \left(\frac{\pi }{4}-\frac{y}{2}\right) \\ \Rightarrow y=2\left(\frac{\pi }{4}-{\tan }^{-1} \left(e^x\right)\right)& \end{array} \end{gathered}$$$$\begin{gathered} \frac{dy}{dx}=2\cdot \frac{-1}{1+{\left(e^x\right)}^2}\cdot e^x \\ \text{ At }t=\frac{1}{2},\tan \frac{y}{2}=\sqrt{\frac{1-\frac{1}{2}}{1+\frac{1}{2}}}=\frac{1}{\sqrt{3}}\Rightarrow \frac{y}{2}=\frac{\pi }{6} \\ \Rightarrow y=\frac{\pi }{3}, e^x=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3} \end{gathered}$$$\begin{array}{rr}\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2(2-\sqrt{3})}{1+(2-\sqrt{3}{)}^2}=\frac{-2(2-\sqrt{3})}{8-4\sqrt{3}}=-\frac{2}{4}=-\frac{1}{2}& \mathrm{ }\end{array}$