Answer
$$ f'(x ) =\frac{-2\cos x}{(\sin x -1)^2} .$$
Work Step by Step
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$
Recall that $(\sin x)'=\cos x$.
Recall that $(\cos x)'=-\sin x$.
Since $ f(x )= \frac{\sin x +1 }{\sin x -1}$, then using the quotient rule, the derivative is given by $$ f'(x )=\frac{(\sin x -1)(\sin x +1)'- (\sin x +1)(\sin x -1)'}{(\sin x -1)^2}\\
=\frac{(\sin x -1)(\cos x)- (\sin x +1)(\cos x)}{(\sin x -1)^2}\\
=\frac{\sin x\cos x -\cos x- \sin x\cos x -\cos x}{(\sin x -1)^2}
\\ =\frac{-2\cos x}{(\sin x -1)^2} .$$
