## Proof

We start by defining $\mathrm{tanh}(x)$ as $\frac{\mathrm{sinh}(x)}{\mathrm{cosh}(x)}$. To find the derivative, we use the quotient rule, which states that the derivative of a quotient $\frac{u}{v}$ is $\frac{{u}^{\prime}v-u{v}^{\prime}}{{v}^{2}}$.

Here, let $u=\mathrm{sinh}(x)$ and $v=\mathrm{cosh}(x)$. The derivatives are ${u}^{\prime}=\mathrm{cosh}(x)$ and ${v}^{\prime}=\mathrm{sinh}(x)$.

Applying the quotient rule, we have:

$$\frac{\text{d}}{\text{d}x}\mathrm{tanh}(x)=\frac{\mathrm{cosh}(x)\mathrm{cosh}(x)-\mathrm{sinh}(x)\mathrm{sinh}(x)}{{\mathrm{cosh}}^{2}(x)}$$This simplifies to:

$$\frac{{\mathrm{cosh}}^{2}(x)-{\mathrm{sinh}}^{2}(x)}{{\mathrm{cosh}}^{2}(x)}$$Using the identity ${\mathrm{cosh}}^{2}(x)-{\mathrm{sinh}}^{2}(x)=1$, we get:

$$\frac{1}{{\mathrm{cosh}}^{2}(x)}={\text{sech}}^{2}(x)$$Thus, the derivative of $\mathrm{tanh}(x)$ is:

$$\overline{){\text{sech}}^{2}(x)}$$## Explanation

To understand the derivative of $\mathrm{tanh}(x)$, let’s start by recognizing that $\mathrm{tanh}(x)$ is defined as the ratio of the hyperbolic sine function $\mathrm{sinh}(x)$ to the hyperbolic cosine function $\mathrm{cosh}(x)$, so $\mathrm{tanh}(x)=\frac{\mathrm{sinh}(x)}{\mathrm{cosh}(x)}$. This represents the hyperbolic tangent, which is commonly used in calculus and hyperbolic geometry.

To find the derivative of $\mathrm{tanh}(x)$, we apply the <strong>quotient rule.</strong> The quotient rule helps us differentiate functions that are expressed as the quotient of two other functions. Specifically, if a function is given as $\frac{u}{v}$, its derivative is $\frac{{u}^{\prime}v-u{v}^{\prime}}{{v}^{2}}$, where $u$ and $v$ are differentiable functions of $x$.

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

Applying the quotient rule, we substitute the derivatives:

$$\frac{\text{d}}{\text{d}x}\mathrm{tanh}(x)=\frac{\mathrm{cosh}(x)\mathrm{cosh}(x)-\mathrm{sinh}(x)\mathrm{sinh}(x)}{{\mathrm{cosh}}^{2}(x)}$$The numerator simplifies to ${\mathrm{cosh}}^{2}(x)-{\mathrm{sinh}}^{2}(x)$. According to the hyperbolic identity, ${\mathrm{cosh}}^{2}(x)-{\mathrm{sinh}}^{2}(x)=1$. This simplifies the entire expression to:

$$\frac{1}{{\mathrm{cosh}}^{2}(x)}$$The expression $\frac{1}{{\mathrm{cosh}}^{2}(x)}$ is the definition of ${\text{sech}}^{2}(x)$, which represents the square of the hyperbolic secant function.

Therefore, the derivative of $\mathrm{tanh}(x)$ with respect to $x$ is $\overline{){\text{sech}}^{2}(x)}$.

*Q.E.D.*