Differentiation of Inverse Hyperbolic Functions

Neetesh Kumar | February 08, 2025 $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space$ Share this Page on:

• 1. Definition and Formula of Inverse Hyperbolic Derivatives
• 2. Explanation & Intuition
• 3. Solved Examples
• 4. Conclusion

The inverse hyperbolic functions are the inverse functions of the hyperbolic functions. The derivatives of these inverse functions are as follows:

1. Derivative of $\text{arsinh}(x)$:

   $\dfrac{d}{dx} \bigg(\text{arsinh}(x)\bigg) = \dfrac{1}{\sqrt{x^2 + 1}}$

2. Derivative of $\text{arcosh}(x)$:

   $\frac{d}{dx} \bigg(\text{arcosh}(x)\bigg) = \dfrac{1}{\sqrt{x^2 - 1}}$

3. Derivative of $\text{artanh}(x)$:

   $\frac{d}{dx} \bigg(\text{artanh}(x)\bigg) = \dfrac{1}{1 - x^2}$

4. Derivative of $\text{arcoth}(x)$:

   $\frac{d}{dx} \bigg(\text{arcoth}(x)\bigg) = \dfrac{1}{1 - x^2}$

5. Derivative of $\text{arsech}(x)$:

   $\frac{d}{dx} \bigg(\text{arsech}(x)\bigg) = -\dfrac{1}{x \sqrt{1 - x^2}}$

6. Derivative of $\text{arcsch}(x)$:

   $\frac{d}{dx} \bigg(\text{arcsch}(x)\bigg) = -\dfrac{1}{|x| \sqrt{1 + x^2}}$

Example 1: Differentiating the Inverse Hyperbolic Sine Function

Differentiate $f(x) = \text{arsinh}(3x)$.

Solution:

Using the derivative of the inverse hyperbolic sine function, we know:

$\dfrac{d}{dx} \bigg(\text{arsinh}(3x)\bigg) = \dfrac{1}{\sqrt{(3x)^2 + 1}} \cdot \dfrac{d}{dx} (3x)$

The derivative of $3x$ is $3$, so:

$f'(x) = \dfrac{3}{\sqrt{9x^2 + 1}}$

Thus, the final answer is: $\boxed{\dfrac{3}{\sqrt{9x^2 + 1}}}$.

Example 2: Differentiating the Inverse Hyperbolic Cosine Function

Differentiate $f(x) = \text{arcosh}(2x + 1)$.

Solution:

Using the derivative of the inverse hyperbolic cosine function, we know:

$\dfrac{d}{dx} \bigg(\text{arcosh}(2x + 1)\bigg) = \dfrac{1}{\sqrt{(2x + 1)^2 - 1}} \cdot \dfrac{d}{dx} (2x + 1)$

The derivative of $2x + 1$ is $2$, so:

$f'(x) = \dfrac{2}{\sqrt{(2x + 1)^2 - 1}}$

Simplifying the denominator:

$f'(x) = \dfrac{2}{\sqrt{4x^2 + 4x}}$

Thus, the final answer is: $\boxed{\dfrac{2}{\sqrt{4x^2 + 4x}}}$.

Example 3: Differentiating the Inverse Hyperbolic Tangent Function

Differentiate $f(x) = \text{artanh}(x^2 + 1)$.

Solution:

Using the derivative of the inverse hyperbolic tangent function, we know:

$\dfrac{d}{dx} \bigg(\text{artanh}(x^2 + 1)\bigg) = \bigg(\dfrac{1}{1 - (x^2 + 1)^2}\bigg) \cdot \dfrac{d}{dx} (x^2 + 1)$

The derivative of $x^2 + 1$ is $2x$, so:

$f'(x) = \dfrac{2x}{1 - (x^2 + 1)^2}$

Simplifying:

$f'(x) = \dfrac{2x}{1 - (x^4 + 2x^2 + 1)}$

Thus, the final answer is: $\boxed{\dfrac{2x}{-x^4 - 2x^2}}$.

Example 4: Differentiating the Inverse Hyperbolic Secant Function

Differentiate $f(x) = \text{arsech}(3x + 2)$.

Solution:

Using the derivative of the inverse hyperbolic secant function, we know:

$\frac{d}{dx} \text{arsech}(3x + 2) = \bigg(-\dfrac{1}{(3x + 2)\sqrt{1 - (3x + 2)^2}}\bigg) \cdot \dfrac{d}{dx} (3x + 2)$

The derivative of $3x + 2$ is $3$, so:

$f'(x) = -\dfrac{3}{(3x + 2)\sqrt{1 - (3x + 2)^2}}$

Thus, the final answer is: $\boxed{-\frac{3}{(3x + 2)\sqrt{1 - (3x + 2)^2}}}$.