Calculus of a Single Variable, Chapter 2, 2.3, Section 2.3, Problem 47

You need to evaluate the derivative of the given function, using the quotient rule, such that:
y' = ((3 - 3sin x)'*(2 cos x) - (3 - 3 sin x)*(2 cos x)')/(4 cos^2 x)
y' = (-3cos x*(2 cos x) + 2 sin x*(3 - 3 sin x))/(4 cos^2 x)
y' = (-6 cos^2 x+ 6sin x - 6 sin^2 x)/(4 cos^2 x)
Factoring out -6 yields:
y' = (-6 (cos^2 x - sin x + sin^2 x))/(4 cos^2 x)
Using sin^2 x + cos^2 x = 1 yields:
y' = (-6/4)*(1 - sin x)/(cos^2 x)
y' = (-3/2)*(1 - sin x)/(cos^2 x)
Hence, evaluating the derivative of the function yields
y' = (-3/2)*(1 - sin x)/(cos^2 x).