$y=\ln\sqrt{ \dfrac{1+\sin x}{1-\sin x}}$
$=\ln\left( \dfrac{1+\sin x}{1-\sin x}\right)^{\frac{1}{2}}$
$=\dfrac{1}{2}\ln \dfrac{1+\sin x}{1-\sin x}$
$y'=\dfrac{1}{2}.\dfrac{1}{ \dfrac{1+\sin x}{1-\sin x}}.\left( \dfrac{1+\sin x}{1-\sin x}\right)'$
$=\dfrac{1}{2}.\dfrac{1-\sin x}{1+\sin x}.\dfrac{ \cos x(1-\sin x)+\cos x(1+\sin x)}{(1-\sin x)^2}$
$=\dfrac{1}{2}.\dfrac{1}{(1+\sin x)(1-\sin x)}.2\cos x$
$=\dfrac{\cos x}{1-\sin^2x}$
$=\dfrac{\cos x}{\cos^2x}$
$=\dfrac{1}{\cos x}$
Vậy $y'=\dfrac{1}{\cos x}$
