## Calculus (3rd Edition)

$$y'=\frac{ \cos t }{\sin t +1}.$$
Recall that $(\ln x)'=\dfrac{1}{x}$ Recall that $(\sin x)'=\cos x$. Since $y=\ln (\sin t +1 )$, then we have $$y'=\frac{ \cos t }{\sin t +1}.$$