Certain combinations of the exponential functions $e^x$ and $e^{-x}$ occur
frequently in mathematics, science, and engineering. These are called
hyperbolic functions.
Definitions of hyperbolic functions:
$$ \sinh x = \frac{e^{x} - e^{-x}}{2} $$
$$ \cosh x = \frac{e^{x} + e^{-x}}{2} $$
$$ \tanh x = \frac{\sinh x}{\cosh x} $$
$$ coth x = \frac{\cosh x}{\sinh x} $$
$$ sech x = \frac{1}{\cosh x} $$
$$ csch x = \frac{1}{\sinh x} $$
Names are read as “hyperbolic sine,” “hyperbolic cosine,” and so on.
Fundamental Hyperbolic Identities
$$ \cosh^{2} x - \sinh^{2} x = 1 $$
$$ \tanh^{2} x + sech^{2} x = 1 $$
$$ coth^{2} x - csch^{2} x = 1 $$
$$ \sinh 2x = 2 \sinh x \cosh x $$
$$ \cosh 2x = \cosh^{2} x + \sinh^{2} x
= 1 + 2\sinh^{2} x
= 2\cosh^{2} x - 1 $$
Differentiation of Hyperbolic Functions
The derivative rules for hyperbolic functions (where $u$ is a function of $x$) are:
$$ \frac{d}{dx}(\sinh u) = \cosh u \, \frac{du}{dx} $$
$$ \frac{d}{dx}(\cosh u) = \sinh u \, \frac{du}{dx} $$
$$ \frac{d}{dx}(\tanh u) = sech^{2} u \, \frac{du}{dx} $$
$$ \frac{d}{dx}(coth u) = -csch^{2} u \, \frac{du}{dx} $$
$$ \frac{d}{dx}(sech u) = -sech u \tanh u \, \frac{du}{dx} $$
$$ \frac{d}{dx}(csch u) = -csch u \coth u \, \frac{du}{dx} $$
Differentiating sinh with Chain Rule — CE Board
Find $\dfrac{dy}{dx}$ if $y = \sinh(3x)$.
Apply the chain rule with $u = 3x$:
$\dfrac{dy}{dx} = \cosh(3x)\cdot 3$
$\boxed{\dfrac{dy}{dx} = 3\cosh(3x)}$
Differentiating tanh with Chain Rule — CE Board
Find $\dfrac{dy}{dx}$ if $y = \tanh(x^2)$.
Apply the chain rule with $u = x^2$:
$\dfrac{dy}{dx} = \text{sech}^2(x^2)\cdot 2x$
$\boxed{\dfrac{dy}{dx} = 2x\,\text{sech}^2(x^2)}$
Hyperbolic Product Differentiation — CE Board
Find $y'$ if $y = x\cosh x$.
Apply the product rule $(uv)' = u'v + uv'$:
$y' = (1)\cosh x + x(\sinh x)$
$\boxed{y' = \cosh x + x\sinh x}$
Verifying the Fundamental Hyperbolic Identity
Using the definitions $\cosh x = \dfrac{e^x + e^{-x}}{2}$ and $\sinh x = \dfrac{e^x - e^{-x}}{2}$, prove that $\cosh^2 x - \sinh^2 x = 1$.
$\cosh^2 x - \sinh^2 x = \left(\dfrac{e^x+e^{-x}}{2}\right)^2 - \left(\dfrac{e^x-e^{-x}}{2}\right)^2$
Using the difference of squares $A^2 - B^2 = (A+B)(A-B)$:
