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

Solution :
$$\eqalign{ & {\text{Let }}t = \cos \,2\theta \cr & {\text{Then }}{e^x} = \frac{{\sqrt {1 + \cos \,2\theta } - \sqrt {1 - \cos \,2\theta } }}{{\sqrt {1 + \cos \,2\theta } + \sqrt {1 - \cos \,2\theta } }} \cr & = \frac{{\cos \,\theta - \sin \,\theta }}{{\cos \,\theta + \sin \,\theta }} \cr & = \frac{{1 - \tan \,\theta }}{{1 + \tan \,\theta }} \cr & = \tan \left( {\frac{\pi }{4} - \theta } \right) \cr & \tan \frac{y}{2} = \sqrt {\frac{{1 - \cos \,2\theta }}{{1 + \cos \,2\theta }}} = \tan \,\theta \cr & {\text{At }}t = \frac{1}{2},\,\cos \,2\theta = \frac{1}{2}\,\,\,\,\,\,\, \Rightarrow \theta = \frac{\pi }{6} \cr & {\text{Then }}x = \log \,\tan \frac{\pi }{{12}},\,\,y = \frac{\pi }{3} \cr & {\text{Differentiating w}}{\text{.r}}{\text{.t}}{\text{.}}\,\theta ,\,{e^x}{\text{.}}\frac{{dx}}{{d\theta }} = - {\sec ^2}\left( {\frac{\pi }{4} - \theta } \right) \cr & {\text{and }}\frac{1}{2}{\sec ^2}\frac{y}{2}.\frac{{dy}}{{d\theta }} = {\sec ^2}\theta \cr & \therefore \,\,\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}} = \frac{{2{{\sec }^2}\theta .{{\cos }^2}\frac{y}{2}}}{{ - {e^{ - x}}.{{\sec }^2}\left( {\frac{\pi }{4} - \theta } \right)}} \cr & {\text{At }}t = \frac{1}{2},\,{\text{i}}{\text{.e}}{\text{., }}\theta = \frac{\pi }{6},\,\frac{{dy}}{{dx}} = \frac{{2{{\sec }^2}\frac{\pi }{6}.{{\cos }^2}\frac{\pi }{6}}}{{ - {e^{ - \log \,\tan \,\frac{\pi }{{12}}}}.{{\sec }^2}\frac{\pi }{{12}}}} \cr & \therefore \,\frac{{dy}}{{dx}} = \frac{2}{{ - \cot \,\frac{\pi }{{12}}.{{\sec }^2}\frac{\pi }{{12}}}} \cr & = - 2\sin \,\frac{\pi }{{12}}.\cos \frac{\pi }{{12}} \cr & = - \sin \frac{\pi }{6} \cr & = - \frac{1}{2} \cr} $$