x = 3 tan t and y = 3 sec t So that ${{dx} \over {dt}}$ = 3sec 2 t and ${{dy} \over {dt}}$ = 3 sec t tan t ${{dy} \over {dx}}$ = ${{dy/dt} \over {dx/dt}}$ = sin t ${{{d^2}y} \over {d{x^2}}}$ = (cos t)$.{{dt} \over {dx}}$ ${{{d^2}y} \over {d{x^2}}} = \left( {\cos t} \right).{1 \over {3{{\sec }^2}t}}$ ${{{d^2}y} \over {d{x^2}}}$ = ${1 \over 3}$(cos 3 t) ${\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{t = \pi /4}}$ = ${1 \over 3} \times {\left( {{1 \over {\sqrt 2 }}} \right)^3} = {1 \over {6\sqrt 2 }}$