Answer :

[tex]y=ln(cosx)\\\\We\ know:\\(lnx)'=\frac{1}{x}\ and\ (cosx)'=-sinx\ and\ \{f[g(x)]\}'=f'[g(x)]\cdot g'(x)\\\\therefore:\\\\y'=\frac{1}{cosx}\cdot(-sinx)=-\frac{sinx}{cosx}=-tanx[/tex]

Other Questions