- (a) By writing \(\sec \theta = \frac { 1 } { \cos \theta }\), show that \(\frac { \mathrm { d } } { \mathrm { d } \theta } ( \sec \theta ) = \sec \theta \tan \theta\)
(b) Given that
$$x = \mathrm { e } ^ { \sec y } \quad x > \mathrm { e } , \quad 0 < y < \frac { \pi } { 2 }$$
show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { x \sqrt { \mathrm {~g} ( x ) } } , \quad x > \mathrm { e }$$
where \(\mathrm { g } ( x )\) is a function of \(\ln x\).