University Calculus: Early Transcendentals (3rd Edition)

$$g'(t)=-\frac{3\cos3t(3-2t)+2(1-\sin3t)}{(1+\sin3t)^2}$$
$$g(t)=\Big(\frac{1+\sin3t}{3-2t}\Big)^{-1}$$ According to the Chain Rule, we have $$g'(t)=-\Big(\frac{1+\sin3t}{3-2t}\Big)^{-2}\Big(\frac{1+\sin3t}{3-2t}\Big)'$$ - Calculate $\Big(\frac{1+\sin3t}{3-2t}\Big)'$: $$\Big(\frac{1+\sin3t}{3-2t}\Big)'=\frac{(1+\sin3t)'(3-2t)-(1+\sin3t)(3-2t)'}{(3-2t)^2}$$ We have $(1+\sin3t)'=0+(\sin3t)'=\cos3t\times(3t)'=3\cos3t$ and $(3-2t)=-2$ $$\Big(\frac{1+\sin3t}{3-2t}\Big)'=\frac{3\cos3t(3-2t)+2(1+\sin3t)}{(3-2t)^2}$$ Replace back to the calculation of $g'(t)$: $$g'(t)=-\Big(\frac{1+\sin3t}{3-2t}\Big)^{-2}\frac{3\cos3t(3-2t)+2(1+\sin3t)}{(3-2t)^2}$$ $$g'(t)=-\frac{(3-2t)^2}{(1+\sin3t)^2}\frac{3\cos3t(3-2t)+2(1+\sin3t)}{(3-2t)^2}$$ $$g'(t)=-\frac{3\cos3t(3-2t)+2(1-\sin3t)}{(1+\sin3t)^2}$$