1. **State the problem:** Find the derivative $f'(x)$ of the function $f(x) = \csc x \cot x$. 2. **Recall the formulas and rules:** - The derivative of $\csc x$ is $-\csc x \cot x$. - The derivative of $\cot x$ is $-\csc^2 x$. - Use the product rule: If $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$. 3. **Apply the product rule:** Let $u = \csc x$ and $v = \cot x$. Then, $$ f'(x) = u'v + uv' = (-\csc x \cot x)(\cot x) + (\csc x)(-\csc^2 x) $$ 4. **Simplify each term:** $$ (-\csc x \cot x)(\cot x) = -\csc x \cot^2 x $$ $$ (\csc x)(-\csc^2 x) = -\csc^3 x $$ 5. **Combine the terms:** $$ f'(x) = -\csc x \cot^2 x - \csc^3 x $$ 6. **Factor out common terms:** $$ f'(x) = -\csc x (\cot^2 x + \csc^2 x) $$ 7. **Use the Pythagorean identity:** Recall that $\cot^2 x + 1 = \csc^2 x$, so $$ \cot^2 x + \csc^2 x = \cot^2 x + (\cot^2 x + 1) = 2\cot^2 x + 1 $$ Thus, $$ f'(x) = -\csc x (2\cot^2 x + 1) $$ **Final answer:** $$ f'(x) = -\csc x (2\cot^2 x + 1) $$