Answer
$y'=\dfrac{3\sin(x)-3}{2\cos^2(x)}$
Work Step by Step
Using the quotient rule: $y'=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=(1-\sin(x)); u'(x)=-\cos(x)$
$v(x)=\cos(x); v'(x)=-\sin(x)$
$y'=\frac{3}{2}(\frac{(-\cos(x))(cos(x))-(-\sin(x))(1-\sin(x))}{\cos^2(x)})$
$=\frac{3}{2}(\frac{-\cos^2(x)+\sin(x)-\sin^2(x)}{cos^2(x)})$
$=\frac{3}{2}(\frac{-(\cos^2(x)+\sin^2(x))+\sin(x)}{\cos^2(x)})$
$=\frac{3}{2}(\frac{\sin(x)-1}{\cos^2(x)})$
$=\frac{3\sin(x)-3}{2\cos^2(x)}$
