1. **Problem:** Find the derivative of $f(x) = \cosh(x) \coth(x)$ and express the answer entirely in terms of $\sinh(x)$, $\cosh(x)$, and/or $\tanh(x)$. 2. **Recall the product rule:** If $f(x) = u(x)v(x)$, then $$f'(x) = u'(x)v(x) + u(x)v'(x).$$ 3. **Identify functions:** Let $u(x) = \cosh(x)$ and $v(x) = \coth(x)$. 4. **Derivatives:** - $u'(x) = \sinh(x)$ because $\frac{d}{dx} \cosh(x) = \sinh(x)$. - $v(x) = \coth(x) = \frac{\cosh(x)}{\sinh(x)}$. 5. **Find $v'(x)$:** Use quotient rule or known derivative: $$\frac{d}{dx} \coth(x) = -\csch^2(x) = -\frac{1}{\sinh^2(x)}.$$ 6. **Apply product rule:** $$f'(x) = u'(x)v(x) + u(x)v'(x) = \sinh(x) \coth(x) + \cosh(x) \left(-\frac{1}{\sinh^2(x)}\right).$$ 7. **Rewrite $\coth(x)$:** $$\coth(x) = \frac{\cosh(x)}{\sinh(x)}.$$ 8. **Substitute:** $$f'(x) = \sinh(x) \frac{\cosh(x)}{\sinh(x)} - \frac{\cosh(x)}{\sinh^2(x)} = \cosh(x) - \frac{\cosh(x)}{\sinh^2(x)}.$$ 9. **Factor out $\cosh(x)$:** $$f'(x) = \cosh(x) \left(1 - \frac{1}{\sinh^2(x)}\right) = \cosh(x) \frac{\sinh^2(x) - 1}{\sinh^2(x)}.$$ 10. **Use identity:** $$\cosh^2(x) - \sinh^2(x) = 1 \implies \sinh^2(x) - 1 = \cosh^2(x) - 2.$$ 11. **Rewrite numerator:** $$\sinh^2(x) - 1 = (\cosh^2(x) - 1) - 1 = \sinh^2(x) - 1.$$ Actually, better to keep as is or use the identity directly: $$\sinh^2(x) - 1 = \cosh^2(x) - 2$$ is incorrect, so keep as $\sinh^2(x) - 1$. 12. **Final expression:** $$f'(x) = \cosh(x) \frac{\sinh^2(x) - 1}{\sinh^2(x)}.$$ This is fully expressed in terms of $\sinh(x)$ and $\cosh(x)$, and since $\tanh(x) = \frac{\sinh(x)}{\cosh(x)}$, this satisfies the requirement. **Answer:** $$\boxed{\frac{d}{dx} \left( \cosh(x) \coth(x) \right) = \cosh(x) \frac{\sinh^2(x) - 1}{\sinh^2(x)}}.$$