1. We are asked to find the derivative of the function $$y = \frac{1}{x \sin x}$$. 2. This function is a quotient, so we can rewrite it as $$y = (x \sin x)^{-1}$$ to use the chain rule and product rule. 3. Let $$u = x \sin x$$, then $$y = u^{-1}$$. 4. The derivative of $$y$$ with respect to $$x$$ is $$y' = -1 \cdot u^{-2} \cdot u' = -\frac{u'}{u^2}$$. 5. Now we find $$u'$$. Since $$u = x \sin x$$, use the product rule: $$u' = (x)' \sin x + x (\sin x)' = 1 \cdot \sin x + x \cos x = \sin x + x \cos x$$. 6. Substitute back into the derivative formula: $$y' = -\frac{\sin x + x \cos x}{(x \sin x)^2}$$. 7. This is the derivative of the given function. Final answer: $$y' = -\frac{\sin x + x \cos x}{x^2 \sin^2 x}$$