Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 22

Find $y'$ of $y = \displaystyle \frac{1}{\sin(x- \sin x)}$

$
\begin{equation}
\begin{aligned}
y' &= \frac{d}{dx} \left( \frac{1}{\sin(x- \sin x)} \right) \\
\\
y' &= \frac{[\sin(x-\sin x)](0) - [\cos(x-\sin x)] \frac{d}{dx} ( x - \sin x) }{\sin^2(x-\sin x)}\\
\\
y' &= \frac{[\sin(x - \sin x)](0) - [\cos ( x - \sin x) ] \frac{d}{dx}(x - \sin x) }{\sin^2(x-\sin x)}\\
\\
y' &= \frac{\left[-\cos (x - \sin x )\right] (1-\cos x)}{\sin^2(x-\sin x)}\\
\\
y' &= - \left( \frac{\cos ( x - \sin x)}{\sin^2(x-\sin x)} \right) \left( \frac{1}{\sin(x-\sin x)} \right) ( 1 - \cos x) \qquad \text{Note that } \frac{\cos x}{\sin x } = \cot x \text{ and } \frac{1}{\sin x} = \csc x\\
\\
y' &= -\cot(x - \sin x) \csc ( x - \sin x) ( 1 - \cos x) \qquad \text{or} \qquad y' = \cot ( x - \sin x ) \csc ( x - \sin x ) ( \cos x + 1)
\end{aligned}
\end{equation}
$