Hyperbolic Functions
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:
$$ \frac{d}{dx}(\sinh u) = \cosh u \, \frac{du}{dx} $$
$$ \frac{d}{dx}(\cosh u) = \sinh u \, \frac{du}{dx} $$
$$ \frac{d}{dx}(\tanh u) = sech^{2} u \, \frac{du}{dx} $$
$$ \frac{d}{dx}(coth u) = -csch^{2} u \, \frac{du}{dx} $$
$$ \frac{d}{dx}(sech u) = -sech u \tanh u \, \frac{du}{dx} $$
$$ \frac{d}{dx}(csch u) = -csch u \coth u \, \frac{du}{dx} $$
Proofs for sinh and tanh Derivatives
Proof of $\dfrac{d}{dx}(\sinh u)$. By definition,
$$ \sinh u = \frac{e^u - e^{-u}}{2} $$
Differentiating with respect to $x$:
$$ \frac{d}{dx}(\sinh u)
= \frac{d}{dx}\left(\frac{e^u - e^{-u}}{2}\right) $$
$$ = \frac{1}{2}\left(e^u\frac{du}{dx} - e^{-u}\left(-\frac{du}{dx}\right)\right) $$
$$ = \left(\frac{e^u + e^{-u}}{2}\right)\frac{du}{dx} $$
$$ \boxed{\frac{d}{dx}(\sinh u) = \cosh u \frac{du}{dx}} $$
Proof of $\dfrac{d}{dx}(\tanh u)$. By definition,
$$ \tanh u = \frac{\sinh u}{\cosh u} $$
Differentiating with respect to $x$ and applying the quotient rule:
$$ \frac{d}{dx}(\tanh u)
= \frac{\cosh u \frac{d}{dx}(\sinh u) - \sinh u \frac{d}{dx}(\cosh u)}{\cosh^2 u} $$
$$ = \frac{\cosh u\left(\cosh u\frac{du}{dx}\right) - \sinh u\left(\sinh u\frac{du}{dx}\right)}{\cosh^2 u} $$
$$ = \frac{\cosh^2 u - \sinh^2 u}{\cosh^2 u}\frac{du}{dx} $$
$$ = \frac{1}{\cosh^2 u}\frac{du}{dx} $$
$$ \boxed{\frac{d}{dx}(\tanh u) = \operatorname{sech}^2 u \frac{du}{dx}} $$
Board-exam reminder: the chain rule factor $\dfrac{du}{dx}$ is part of every hyperbolic derivative rule when the argument is not simply $x$.
Book Examples on Hyperbolic Differentiation
Example 1. If $y = \sinh(4x + 3)$, find $\dfrac{dy}{dx}$.
Show Solution
Use $\dfrac{d}{dx}(\sinh u) = \cosh u\dfrac{du}{dx}$ with $u = 4x + 3$.
$$ \frac{dy}{dx} = \cosh(4x + 3)\frac{d}{dx}(4x + 3) $$
$$ = 4\cosh(4x + 3) $$
$$ \boxed{\frac{dy}{dx} = 4\cosh(4x + 3)} $$
Example 2. If $y = 3\cosh^2 4x$, find $\dfrac{dy}{dx}$.
Show Solution
Read $\cosh^2 4x$ as $\left(\cosh 4x\right)^2$.
$$ \frac{dy}{dx} = 3\cdot 2\cosh 4x \frac{d}{dx}(\cosh 4x) $$
$$ = 6\cosh 4x(\sinh 4x)(4) $$
$$ = 24\sinh 4x\cosh 4x $$
Using $\sinh 2a = 2\sinh a\cosh a$ with $a = 4x$:
$$ 24\sinh 4x\cosh 4x = 12(2\sinh 4x\cosh 4x) = 12\sinh 8x $$
$$ \boxed{\frac{dy}{dx} = 12\sinh 8x} $$
Exercise 4.8 - Hyperbolic Differentiation
Find $\dfrac{dy}{dx}$ or $y'$ and simplify.
Use the identities $\cosh^2 x - \sinh^2 x = 1$, $1 - \tanh^2 x = \operatorname{sech}^2 x$, and $\sinh 2x = 2\sinh x\cosh x$ when they make the answer shorter.
Derivative of $y = \sinh^2 5x$
Find $y'$ if $y = \sinh^2 5x$.
Show Solution
Write the function as $y = \left(\sinh 5x\right)^2$ and apply the chain rule twice.
$$ y' = 2\sinh 5x \frac{d}{dx}(\sinh 5x) $$
$$ = 2\sinh 5x(\cosh 5x)(5) $$
$$ = 10\sinh 5x\cosh 5x $$
Since $\sinh 10x = 2\sinh 5x\cosh 5x$,
$$ \boxed{y' = 5\sinh 10x} $$
Derivative of $y = \dfrac{\sinh x}{1 + \cosh x}$
Find $y'$ if $y = \dfrac{\sinh x}{1 + \cosh x}$.
Show Solution
Use the quotient rule with numerator $N = \sinh x$ and denominator $D = 1 + \cosh x$.
$$ y' = \frac{(1+\cosh x)(\cosh x) - (\sinh x)(\sinh x)}{(1+\cosh x)^2} $$
$$ = \frac{\cosh x + \cosh^2 x - \sinh^2 x}{(1+\cosh x)^2} $$
Use $\cosh^2 x - \sinh^2 x = 1$.
$$ y' = \frac{\cosh x + 1}{(1+\cosh x)^2}
= \frac{1}{1+\cosh x} $$
$$ \boxed{y' = \frac{1}{1+\cosh x}} $$
Derivative of $y = \ln(\tanh 2x)$
Find $y'$ if $y = \ln(\tanh 2x)$.
Show Solution
Use $\dfrac{d}{dx}[\ln v] = \dfrac{v'}{v}$ with $v = \tanh 2x$.
$$ y' = \frac{1}{\tanh 2x}\frac{d}{dx}(\tanh 2x) $$
$$ = \frac{1}{\tanh 2x}\left(2\operatorname{sech}^2 2x\right) $$
$$ = \frac{2\operatorname{sech}^2 2x}{\tanh 2x} $$
Convert to $\sinh$ and $\cosh$ to simplify.
$$ \frac{2\operatorname{sech}^2 2x}{\tanh 2x}
= \frac{2}{\sinh 2x\cosh 2x}
= \frac{4}{2\sinh 2x\cosh 2x} $$
$$ \boxed{y' = \frac{4}{\sinh 4x} = 4\operatorname{csch}4x} $$
Derivative of $y = \dfrac{x}{2}[\cosh(\ln x) + \sinh(\ln x)]$
Find $y'$ if $y = \dfrac{x}{2}\left[\cosh(\ln x) + \sinh(\ln x)\right]$.
Show Solution
For real $\ln x$, assume $x > 0$. Use the identity $\cosh u + \sinh u = e^u$.
$$ \cosh(\ln x) + \sinh(\ln x) = e^{\ln x} = x $$
$$ y = \frac{x}{2}(x) = \frac{x^2}{2} $$
$$ \boxed{y' = x} $$
Derivative of $y = \cosh^2 6x + \dfrac{1}{2}\cosh 12x$
Find $y'$ if $y = \cosh^2 6x + \dfrac{1}{2}\cosh 12x$.
Show Solution
Differentiate each term separately.
$$ \frac{d}{dx}(\cosh^2 6x)
= 2\cosh 6x(\sinh 6x)(6)
= 12\sinh 6x\cosh 6x $$
$$ \frac{d}{dx}\left(\frac{1}{2}\cosh 12x\right)
= \frac{1}{2}(\sinh 12x)(12)
= 6\sinh 12x $$
Use $\sinh 12x = 2\sinh 6x\cosh 6x$ on the first derivative.
$$ 12\sinh 6x\cosh 6x = 6\sinh 12x $$
$$ y' = 6\sinh 12x + 6\sinh 12x $$
$$ \boxed{y' = 12\sinh 12x} $$
Derivative of $y = \arctan(\sinh x)$
Find $y'$ if $y = \arctan(\sinh x)$.
Show Solution
Use $\dfrac{d}{dx}(\arctan v) = \dfrac{v'}{1+v^2}$ with $v = \sinh x$.
$$ y' = \frac{\cosh x}{1+\sinh^2 x} $$
Since $\cosh^2 x - \sinh^2 x = 1$, then $1+\sinh^2 x = \cosh^2 x$.
$$ y' = \frac{\cosh x}{\cosh^2 x}
= \frac{1}{\cosh x} $$
$$ \boxed{y' = \operatorname{sech}x} $$
Derivative of $y = \arcsin(\tanh 4x)$
Find $y'$ if $y = \arcsin(\tanh 4x)$.
Show Solution
Use $\dfrac{d}{dx}(\arcsin v) = \dfrac{v'}{\sqrt{1-v^2}}$ with $v = \tanh 4x$.
$$ y' = \frac{4\operatorname{sech}^2 4x}{\sqrt{1-\tanh^2 4x}} $$
Use $1-\tanh^2 u = \operatorname{sech}^2 u$. Since $\operatorname{sech}4x > 0$ for real $x$, $\sqrt{\operatorname{sech}^2 4x} = \operatorname{sech}4x$.
$$ y' = \frac{4\operatorname{sech}^2 4x}{\operatorname{sech}4x} $$
$$ \boxed{y' = 4\operatorname{sech}4x} $$
Derivative of $y = e^x\ln(\sinh x)$
Find $y'$ if $y = e^x\ln(\sinh x)$.
Show Solution
For real $\ln(\sinh x)$, the expression requires $\sinh x > 0$, so $x > 0$. Apply the product rule.
$$ y' = e^x\ln(\sinh x) + e^x\frac{d}{dx}[\ln(\sinh x)] $$
$$ \frac{d}{dx}[\ln(\sinh x)] = \frac{\cosh x}{\sinh x} = \coth x $$
$$ y' = e^x\ln(\sinh x) + e^x\coth x $$
$$ \boxed{y' = e^x[\ln(\sinh x) + \coth x]} $$
Differentiating sinh with Chain Rule — CE Board
Find $\dfrac{dy}{dx}$ if $y = \sinh(3x)$.
Show Solution
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 $y = \tanh(x^2)$.
Show Solution
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 $y = x\cosh x$.
Show Solution
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 $\cosh^2 x - \sinh^2 x = 1$.
Show Solution
$\cosh^2 x - \sinh^2 x = \left(\dfrac{e^x+e^{-x}}{2}\right)^2 - \left(\dfrac{e^x-e^{-x}}{2}\right)^2$
Using the difference of squares $A^2 - B^2 = (A+B)(A-B)$:
$= \dfrac{(e^x+e^{-x}+e^x-e^{-x})(e^x+e^{-x}-e^x+e^{-x})}{4} = \dfrac{(2e^x)(2e^{-x})}{4} = \dfrac{4}{4} = \boxed{1}$
Hyperbolic Sum Identity — CE Board
Show that $\sinh x + \cosh x = e^x$, then use it to find the derivative of $f(x) = \sinh x + \cosh x$ at $x = 0$.
Show Solution
Identity:
$\sinh x + \cosh x = \dfrac{e^x - e^{-x}}{2} + \dfrac{e^x + e^{-x}}{2} = \dfrac{2e^x}{2} = e^x$
Derivative: Since $f(x) = e^x$,
$f'(x) = e^x \implies f'(0) = e^0 = \boxed{1}$
Alternatively: $f'(x) = \cosh x + \sinh x$, so $f'(0) = \cosh(0) + \sinh(0) = 1 + 0 = 1$ ✓
