x = $\sqrt {{2^{\cos e{c^{ - 1}}t}}}$ $\therefore\,\,\,\,$ ${{dx} \over {dt}}$ = ${1 \over {2\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$ $\times$ (${2^{\cos e{c^{ - 1}}t}}\,.\,\log 2$) $\times$ ${{ - 1} \over {t\sqrt {{t^2} - 1} }}$ ${{dy} \over {dt}}$ = ${1 \over {2\sqrt {{2^{{{\sec }^{ - 1}}t}}} }}$ $\times$ $\left( {{2^{{{\sec }^{ - 1}}t}}\log 2} \right)$ $\times$ ${1 \over {t\sqrt {{t^2} - 1} }}$ $\therefore\,\,\,\,$ ${{dy} \over {dx}}$ = ${{{{dy} \over {dt}}} \over {{{dx} \over {dt}}}}$ = ${{ - \sqrt {{2^{\cos e{c^{ - 1}}t}}} } \over {\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$ $\times$ ${{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}$ = $- \sqrt {{{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}}$ = $-$ ${y \over x}$