Answer
$-\dfrac{3\cos3t(3-2t)+2(1-\sin3t)}{(1+\sin3t)^2}$
Work Step by Step
Apply the Chain Rule, we get
$g'(t)=-(\dfrac{1+\sin3t}{3-2t})^{-2}(\dfrac{1+\sin3t}{3-2t})'$ ...(1)
Now, consider $(\dfrac{1+\sin3t}{3-2t})'=\dfrac{(1+\sin3t)'(3-2t)-(1+\sin3t)(3-2t)'}{(3-2t)^2}$ ;
or, $(\dfrac{1+\sin3t}{3-2t})'=\dfrac{3\cos3t(3-2t)+2(1+\sin3t)}{(3-2t)^2}$
Equation (1), becomes: $g'(t)=-(\dfrac{1+\sin3t}{3-2t})^{-2}[\dfrac{3\cos3t(3-2t)+2(1+\sin3t)}{(3-2t)^2}]$
Thus, $g'(t)=-(\dfrac{(3-2t)^2}{(1+\sin3t)^2})[\dfrac{3\cos3t(3-2t)+2(1+\sin3t)}{(3-2t)^2}]=-\dfrac{3\cos3t(3-2t)+2(1-\sin3t)}{(1+\sin3t)^2}$
