Differentiation of 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 of Hyperbolic Functions
• 2. Explanation & Properties
• 3. Solved Examples
• 4. Conclusion

Derivatives:

• $\dfrac{d}{dx} \big(\sinh x\big) = \cosh x = \dfrac{e^x + e^{-x}}{2}$

  
  

• $\dfrac{d}{dx} \big(\cosh x\big) = \sinh x = \dfrac{e^x - e^{-x}}{2}$

  
  

• $\dfrac{d}{dx} \big(\tanh x\big) = \text{sech}^2 x$

Example 1: Differentiating a Hyperbolic Function

Differentiate $f(x) = \sinh(3x^2)$.

Solution:

Using the chain rule, let:

• Outer function: $f(u) = \sinh u$, so $f'(u) = \cosh u$.
• Inner function: $g(x) = 3x^2$, so $g'(x) = 6x$.

Applying the chain rule:

$f'(x) = \cosh(3x^2) \cdot 6x$

Thus, the final answer is: $\boxed{6x \cosh(3x^2)}$.

Example 2: Finding the Derivative of a Combination

Differentiate $f(x) = x \cosh x + \sinh x$.

Solution:

Using the product rule for $x \cosh x$ and the basic derivative of $\sinh x$:

$f'(x) = \dfrac{d}{dx} (x. \cosh x) + \dfrac{d}{dx} (\sinh x)$

Using the product rule:

$f'(x) = x. \sinh x + \cosh x + \cosh x$

$f'(x) = x. \sinh x + 2.\cosh x$

Thus, the final answer is: $\boxed{x. \sinh x + 2.\cosh x}$.

Example 3: Hyperbolic Tangent in Electrical Engineering

In circuit analysis, the function $I(t) = \tanh(2t)$ models current flow. Find $I'(t)$.

Solution:

Using the derivative of hyperbolic tangent:

$I'(t) = \dfrac{d}{dt} \big(\tanh(2t)\big)$

Since $\dfrac{d}{dx} \big(\tanh x\big) = \text{sech}^2 x$, and using the chain rule:

$I'(t) = \text{sech}^2 (2t) \cdot 2$

Thus, the final answer is: $\boxed{2 \text{sech}^2 (2t)}$.

Example 4: Hyperbolic Function in Relativity

The velocity of an object in special relativity is modeled by: $v(t) = c \tanh\bigg(\dfrac{at}{c}\bigg)$,

where $c$ is the speed of light, and $a$ is acceleration. Find $v'(t)$.

Solution:

Using the chain rule:

$v'(t) = c \cdot \dfrac{d}{dt} \bigg(\tanh \left(\frac{at}{c}\right)\bigg)$

Using the derivative of $\tanh x$:

$v'(t) = c \cdot \text{sech}^2 \left(\dfrac{at}{c}\right) \cdot \dfrac{a}{c}$

Simplifying:

$v'(t) = a \text{sech}^2 \left(\dfrac{at}{c}\right)$

Thus, the final answer is: $\boxed{a \text{sech}^2 \left(\dfrac{at}{c}\right)}$.