Chain rule with cosecant of a reciprocal function

Identify the outer function as $f(u) = \csc u$ and the inner function as $u = g(x) = \dfrac{1}{x} = x^{-1}$. Since $f'(u) = -\csc u \cot u$, the chain rule gives $\dfrac{d}{dx}\left[\csc(u)\right] = -\csc(u)\cot(u) \cdot u'$.

Now differentiate the inner function: $u' = \dfrac{d}{dx}\left[x^{-1}\right] = -x^{-2} = -\dfrac{1}{x^2}$. Substituting back, $g'(x) = -\csc\left(\dfrac{1}{x}\right)\cot\left(\dfrac{1}{x}\right) \cdot \left(-\dfrac{1}{x^2}\right) = \dfrac{1}{x^2}\csc\left(\dfrac{1}{x}\right)\cot\left(\dfrac{1}{x}\right)$.