Derivative of tanh(x) - Proof and Explanation

Proof

We start by defining $\tanh (x)$ as $\frac{\sinh (x)}{\cosh (x)}$. 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 $u=\sinh (x)$ and $v=\cosh (x)$. The derivatives are $u^{\prime }=\cosh (x)$ and $v^{\prime }=\sinh (x)$.

Applying the quotient rule, we have:

This simplifies to:

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

Thus, the derivative of $\tanh (x)$ is:

Explanation

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

To find the derivative of $\tanh (x)$, 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 $\tanh (x)$, we have $u=\sinh (x)$ and $v=\cosh (x)$. The derivative of $\sinh (x)$ is $\cosh (x)$, and the derivative of $\cosh (x)$ is $\sinh (x)$.

Applying the quotient rule, we substitute the derivatives:

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

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

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

Q.E.D.