Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when
$$y = \mathrm { e } ^ { 3 x } + \ln x$$
Given that \(u = \frac { \sin x } { 1 + \cos x }\), show that \(\frac { \mathrm { d } u } { \mathrm {~d} x } = \frac { 1 } { 1 + \cos x }\).
Hence show that if \(y = \ln \left( \frac { \sin x } { 1 + \cos x } \right)\), then \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \operatorname { cosec } x\).