If [x = 2^ ec^ - 1 ] and [y = 2^sec^ - 1t ,,( | t | 1 ),] then [dy dx] is equal to :

Question

If $$x = \sqrt {{2^{\cos e{c^{ - 1}}}}} $$ and $$y = \sqrt {{2^{se{c^{ - 1}}t}}} \,\,\left( {\left| t \right| \ge 1} \right),$$ then $${{dy} \over {dx}}$$ is equal to :

Accepted Answer

x =  $$\sqrt {{2^{\cos e{c^{ - 1}}t}}} $$

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

$${{dy} \over {dt}}$$ = $${1 \over {2\sqrt {{2^{{{\sec }^{ - 1}}t}}} }}$$ $$ \times $$ $$\left( {{2^{{{\sec }^{ - 1}}t}}\log 2} \right)$$ $$ \times $$ $${1 \over {t\sqrt {{t^2} - 1} }}$$

$$\therefore\,\,\,\,$$ $${{dy} \over {dx}}$$

= $${{{{dy} \over {dt}}} \over {{{dx} \over {dt}}}}$$

= $${{ - \sqrt {{2^{\cos e{c^{ - 1}}t}}} } \over {\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$$ $$ \times $$ $${{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}$$

= $$ - \sqrt {{{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}} $$

= - $${y \over x}$$

Solutions · 1