Formulas and examples, with detailed solutions, on the derivatives of hyperbolic functions are presented. For definitions and graphs of hyperbolic functions go to Graphs of Hyperbolic Functions
Table of Hyperbolic Functions and Their Derivatives
\( f(x) = \operatorname{csch} x = \dfrac{1}{\sinh x} \)
\( f '(x) = - \operatorname{csch} x \coth x \)
\( f(x) = \operatorname{sech} x = \dfrac{1}{\cosh x} \)
\( f '(x) = - \operatorname{sech} x \tanh x \)
Examples with Solutions
Example 1
Find the derivative of \( f(x) = \sinh (x^2) \)
Solution to Example 1:
Let \( u = x^2 \) and \( y = \sinh u \) and use the chain rule to find the derivative of the given function \( f \) as follows.
\( f '(x) = \dfrac{dy}{du} \dfrac{du}{dx} \)
\( \dfrac{dy}{du} = \cosh u \), see formula above, and \( \dfrac{du}{dx} = 2x \)
\( f '(x) = 2x \cosh u = 2x \cosh (x^2) \)
Substitute \( u = x^2 \) in \( f '(x) \) to obtain
\[ f '(x) = 2x \cosh (x^2) \]
Example 2
Find the derivative of \( f(x) = 2 \sinh x + 4 \cosh x \)
Solution to Example 2:
Let \( g(x) = 2 \sinh x \) and \( h(x) = 4 \cosh x \), function \( f \) is the sum of functions \( g \) and \( h \): \[ f(x) = g(x) + h(x) \]. Use the sum rule, \( f '(x) = g '(x) + h '(x) \), to find the derivative of function \( f \)
\[ f '(x) = 2 \cosh x + 4 \sinh x \]
Example 3
Find the derivative of \( f(x) = \dfrac{\cosh x}{\sinh (x^2)} \)
Solution to Example 3:
Let \( g(x) = \cosh x \) and \( h(x) = \sinh x^2 \), function \( f \) is the quotient of functions \( g \) and \( h \): \( f(x) = \dfrac{g(x)}{h(x)} \). Hence we use the quotient rule
, \[ f '(x) = \dfrac{h(x) g '(x) - g(x) h '(x)}{h(x)^2} \] to find the derivative of function \( f \).
\( g '(x) = \sinh x \)
\( h '(x) = 2x \cosh x^2 \) (see example 2 above)
\[ f '(x) = \dfrac{( \sinh x^2 ) ( \sinh x ) - ( \cosh x ) ( 2x \cosh x^2 )}{( \sinh x^2 )^2} \]
Example 4
Find the derivative of \( f(x) = (\sinh x)^2 \)
Solution to Example 4:
Let \( u = \sinh x \) and \( y = u^2 \), Use the chain rule to find the derivative of function \( f \) as follows.
\( f '(x) = \dfrac{dy}{du} \dfrac{du}{dx} \)
\( \dfrac{dy}{du} = 2u \) and \( \dfrac{du}{dx} = \cosh x \)
\( f '(x) = 2u \cosh x \)
Put \( u = \sinh x \) in \( f '(x) \) obtained above
\[ f '(x) = 2 \sinh x \cosh x \]