Starting from the definitions of sech and tanh in terms of exponentials, prove that
$$1 - \operatorname { sech } ^ { 2 } t = \tanh ^ { 2 } t$$
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The curve \(C\) has parametric equations
$$\mathrm { x } = \frac { 1 } { 2 } \tanh ^ { 2 } \mathrm { t } + \text { Insecht } , \quad \mathrm { y } = 1 + \tanh ^ { 4 } \mathrm { t } , \quad \text { for } t > 0$$
Show that \(\frac { d y } { d x } = - 4 \operatorname { sech } ^ { 2 } t\).
Find the coordinates of the point on \(C\) with \(\frac { d ^ { 2 } y } { d x ^ { 2 } } = - \frac { 9 } { 2 }\), giving your answer in the form \(( a + \ln b , c )\) where \(a , b\) and \(c\) are rational numbers.
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