Derivative of atanh(x) - Proof and Explanation

Proof

We start by defining $y=\text{atanh}(x)$, which means $x=\tanh (y)$.

1. Definition and implicit differentiation:

   $$y=\text{atanh}(x)⟹x=\tanh (y)$$

2. Derivative of $\tanh (y)$:

   $$\frac{\text{d}}{\text{d}y}\tanh (y)={\text{sech}}^2(y)$$

3. Differentiating implicitly with respect to $x$:

   $$\frac{dx}{dy}={\text{sech}}^2(y)$$

4. Reciprocal to find $\frac{dy}{dx}$:

   $$\frac{dy}{dx}=\frac{1}{{\text{sech}}^2(y)}$$

5. Simplifying using the identity $\text{sech}(y)=\frac{1}{\cosh (y)}$:

   $${\text{sech}}^2(y)=\frac{1}{{\cosh }^2(y)}$$

   So,

   $$\frac{dy}{dx}={\cosh }^2(y)$$

6. Express ${\cosh }^2(y)$ in terms of $x$: From the identity ${\cosh }^2(y)-{\sinh }^2(y)=1$ and knowing $\tanh (y)=x$, we have $\sinh (y)=x\cosh (y)$. Thus,

   $${\cosh }^2(y)-x^2{\cosh }^2(y)=1⟹{\cosh }^2(y)(1-x^2)=1⟹{\cosh }^2(y)=\frac{1}{1-x^2}$$

Therefore,

Thus, the derivative of $\text{atanh}(x)$ is:

Explanation

To understand the derivative of $\text{atanh}(x)$, we first define $y=\text{atanh}(x)$, which means that $x$ is the hyperbolic tangent of $y$. This relationship can be written as $x=\tanh (y)$.

The function $\tanh (y)$ is a hyperbolic function, similar to trigonometric functions but for hyperbolic angles. Its derivative with respect to $y$ is ${\text{sech}}^2(y)$, where $\text{sech}(y)$ is the hyperbolic secant, defined as $\frac{1}{\cosh (y)}$.

Using implicit differentiation, we differentiate both sides of $x=\tanh (y)$ with respect to $x$. This gives us:

To find $\frac{dy}{dx}$, we take the reciprocal of $\frac{dx}{dy}$:

Next, we simplify $\frac{1}{{\text{sech}}^2(y)}$. Since $\text{sech}(y)=\frac{1}{\cosh (y)}$, we have ${\text{sech}}^2(y)=\frac{1}{{\cosh }^2(y)}$. Therefore,

To express ${\cosh }^2(y)$ in terms of $x$, we use the identity ${\cosh }^2(y)-{\sinh }^2(y)=1$. Knowing that $\tanh (y)=x$, we get $\sinh (y)=x\cosh (y)$. Substituting this into the identity, we obtain:

Hence, the derivative of $\text{atanh}(x)$ is:

Q.E.D.