Derivative of coth(x) - Proof and Explanation

Proof

To find the derivative of $\coth (x)$ , we start with its definition:

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

Using the quotient rule, which states that the derivative of a quotient $\frac{u}{v}$ is:

\frac{u^{'} v - u v^{'}}{v^{2}}

Let $u = \cosh (x)$ and $v = \sinh (x)$ . The derivatives are $u^{'} = \sinh (x)$ and $v^{'} = \cosh (x)$ .

Applying the quotient rule:

\frac{d}{d x} \coth (x) = \frac{\sinh (x) \cdot \sinh (x) - \cosh (x) \cdot \cosh (x)}{\sinh^{2} (x)}

Simplifying the numerator:

\sinh^{2} (x) - \cosh^{2} (x) = - 1

Thus, the derivative becomes:

\frac{- 1}{\sinh^{2} (x)} = - {c s c h}^{2} (x)

So, the derivative of $\coth (x)$ is:

\frac{d}{d x} \coth (x) = - {c s c h}^{2} (x)

Explanation

To understand the derivative of $\coth (x)$ , we begin by recognizing its definition as the hyperbolic cotangent function, expressed as $\coth (x) = \frac{\cosh (x)}{\sinh (x)}$ . This function represents the ratio of the hyperbolic cosine to the hyperbolic sine.

We use the quotient rule to find the derivative of a quotient of two functions. According to this rule, for a function $\frac{u}{v}$ , the derivative is $\frac{u^{'} v - u v^{'}}{v^{2}}$ , where $u$ and $v$ are both functions of $x$ .

In the case of $\coth (x)$ , we set $u = \cosh (x)$ and $v = \sinh (x)$ . The derivative of $\cosh (x)$ is $\sinh (x)$ , and the derivative of $\sinh (x)$ is $\cosh (x)$ .

Applying these derivatives in the quotient rule, we have:

\frac{d}{d x} \coth (x) = \frac{\sinh (x) \cdot \sinh (x) - \cosh (x) \cdot \cosh (x)}{\sinh^{2} (x)}

The numerator simplifies to $\sinh^{2} (x) - \cosh^{2} (x)$ . According to the identity $\cosh^{2} (x) - \sinh^{2} (x) = 1$ , we know that $\sinh^{2} (x) - \cosh^{2} (x) = - 1$ .

Thus, the expression becomes:

\frac{- 1}{\sinh^{2} (x)}

Since $\frac{1}{\sinh^{2} (x)}$ is ${c s c h}^{2} (x)$ , the final result is:

- {c s c h}^{2} (x)

Q.E.D.