We want to verify if the derivative of \( \coth(x) \) is equal to \( -\csch^2(x) \).

The hyperbolic cotangent function is defined as:

\[
\coth(x) = \frac{\cosh(x)}{\sinh(x)}
\]

To differentiate \( \coth(x) \), we use the quotient rule, which is given by:

\[
\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}
\]

For \( \coth(x) = \frac{\cosh(x)}{\sinh(x)} \), let:

- \( f(x) = \cosh(x) \)
- \( g(x) = \sinh(x) \)

We know that:

\[
\frac{d}{dx} \cosh(x) = \sinh(x)
\]
\[
\frac{d}{dx} \sinh(x) = \cosh(x)
\]

Now applying the quotient rule:

\[
\frac{d}{dx} \left( \coth(x) \right) = \frac{\sinh(x) \cdot \sinh(x) - \cosh(x) \cdot \cosh(x)}{[\sinh(x)]^2}
\]

This simplifies to:

\[
\frac{d}{dx} \left( \coth(x) \right) = \frac{\sinh^2(x) - \cosh^2(x)}{[\sinh(x)]^2}
\]

Using the hyperbolic identity \( \cosh^2(x) - \sinh^2(x) = 1 \), this becomes:

\[
\frac{d}{dx} \left( \coth(x) \right) = -\frac{1}{\sinh^2(x)}
\]

Since \( \csch(x) = \frac{1}{\sinh(x)} \), we can rewrite this as:

\[
\frac{d}{dx} \left( \coth(x) \right) = -\csch^2(x)
\]

Thus, we have verified that:

\[
\frac{d}{dx} \left( \coth(x) \right) = -\csch^2(x)
\]

### Do you want more details or have any questions?

Here are 5 related questions:
1. How do you differentiate other hyperbolic functions like \( \cosh(x) \) and \( \sinh(x) \)?
2. What is the relationship between hyperbolic and trigonometric functions?
3. Can you show how the chain rule applies to hyperbolic functions?
4. How do you integrate \( \csch^2(x) \)?
5. What are the applications of hyperbolic functions in calculus?

**Tip:** Remember that hyperbolic identities

Learn how to verify that the derivative of the hyperbolic cotangent function, coth(x), is equal to -csch^2(x). This problem uses the quotient rule and hyperbolic identities to solve. Suitable for undergraduate calculus students.

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