## Proof

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

$$\mathrm{coth}(x)=\frac{\mathrm{cosh}(x)}{\mathrm{sinh}(x)}$$Using the quotient rule, which states that the derivative of a quotient $\frac{u}{v}$ is:

$$\frac{{u}^{\prime}v-u{v}^{\prime}}{{v}^{2}}$$Let $u=\mathrm{cosh}(x)$ and $v=\mathrm{sinh}(x)$. The derivatives are ${u}^{\prime}=\mathrm{sinh}(x)$ and ${v}^{\prime}=\mathrm{cosh}(x)$.

Applying the quotient rule:

$$\frac{\text{d}}{\text{d}x}\mathrm{coth}(x)=\frac{\mathrm{sinh}(x)\xb7\mathrm{sinh}(x)-\mathrm{cosh}(x)\xb7\mathrm{cosh}(x)}{{\mathrm{sinh}}^{2}(x)}$$Simplifying the numerator:

$${\mathrm{sinh}}^{2}(x)-{\mathrm{cosh}}^{2}(x)=-1$$Thus, the derivative becomes:

$$\frac{-1}{{\mathrm{sinh}}^{2}(x)}=-{\mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}}^{2}(x)$$So, the derivative of $\mathrm{coth}(x)$ is:

$$\frac{\text{d}}{\text{d}x}\mathrm{coth}(x)=\overline{)-{\mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}}^{2}(x)}$$## Explanation

To understand the derivative of $\mathrm{coth}(x)$, we begin by recognizing its definition as the hyperbolic cotangent function, expressed as $\mathrm{coth}(x)=\frac{\mathrm{cosh}(x)}{\mathrm{sinh}(x)}$. 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 $\frac{{u}^{\prime}v-u{v}^{\prime}}{{v}^{2}}$, where $u$ and $v$ are both functions of $x$.

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

Applying these derivatives in the quotient rule, we have:

$$\frac{\text{d}}{\text{d}x}\mathrm{coth}(x)=\frac{\mathrm{sinh}(x)\xb7\mathrm{sinh}(x)-\mathrm{cosh}(x)\xb7\mathrm{cosh}(x)}{{\mathrm{sinh}}^{2}(x)}$$The numerator simplifies to ${\mathrm{sinh}}^{2}(x)-{\mathrm{cosh}}^{2}(x)$. According to the identity ${\mathrm{cosh}}^{2}(x)-{\mathrm{sinh}}^{2}(x)=1$, we know that ${\mathrm{sinh}}^{2}(x)-{\mathrm{cosh}}^{2}(x)=-1$.

Thus, the expression becomes:

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

$$\overline{)-{\mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}}^{2}(x)}$$*Q.E.D.*