Hyperbolic Functions and Their Derivatives

Function	Definition	Derivative
$\sinh x$	${\displaystyle \frac{e^x-e^{-x}}{2}}$	$\cosh x$
$\cosh x$	${\displaystyle \frac{e^x+e^{-x}}{2}}$	$\sinh x$
$\tanh x$	${\displaystyle \frac{\sinh x}{\cosh x}}$	${\text{sech}}^{2}x$
$\coth x$	${\displaystyle \frac{\cosh x}{\sinh x}}$	$-{\text{csch}}^{2}x$
$\text{sech}x$	${\displaystyle \frac{1}{\cosh x}}$	$-\text{sech}x\tanh x$
$\text{csch}x$	${\displaystyle \frac{1}{\sinh x}}$	$-\text{csch}x\coth x$

Definitions

In many applications, exponential functions appear in the combinations $e^x+e^{-x}$ and $e^x-e^{-x}$.

$$\boxed{\cosh x=\frac{e^x+e^{-x}}{2},\sinh x=\frac{e^x-e^{-x}}{2}}$$

These are called the hyperbolic cosine and hyperbolic sine functions.

Memory aid: $\cosh x$ (like $\cos x$) is an even function, so it uses the "plus" combination. $\sinh x$ (like $\sin x$) is an odd function, so it uses the "minus" combination.

$$\cosh (-x)=\cosh x,\sinh (-x)=-\sinh x$$

Special values: $\cosh 0=1$, $\sinh 0=0$.

Both functions are continuous and differentiable everywhere.

The hyperbolic tangent:

$$\boxed{\tanh x=\frac{\sinh x}{\cosh x}=\frac{e^x-e^{-x}}{e^x+e^{-x}}}$$

This function is odd and satisfies $\underset{x\to +\mathrm{\infty }}{lim}\tanh x=1$ and $\underset{x\to -\mathrm{\infty }}{lim}\tanh x=-1$.

Other Hyperbolic Functions

$$\coth x=\frac{\cosh x}{\sinh x},\text{sech}x=\frac{1}{\cosh x},\text{csch}x=\frac{1}{\sinh x}$$

Note: $\coth x$ and $\text{csch}x$ are not defined at $x=0$.

Hyperbolic Identities

$$\boxed{{\cosh }^{2}x-{\sinh }^{2}x=1}$$$$\cosh x-\sinh x=e^{-x},\cosh x+\sinh x=e^x$$$$\boxed{1-{\tanh }^{2}x={\text{sech}}^{2}x=\frac{1}{{\cosh }^{2}x}}$$$${\cosh }^{2}x-1={\text{csch}}^{2}x=\frac{1}{{\sinh }^{2}x}$$$$\boxed{\sinh (x\pm y)=\sinh x\cosh y\pm \cosh x\sinh y}$$$$\boxed{\cosh (x\pm y)=\cosh x\cosh y\pm \sinh x\sinh y}$$

Why the Prefix "Hyper"?

The point $P(\cos \theta ,\sin \theta )$ lies on the unit circle $x^2+y^2=1$, and $\theta$ equals twice the area of the circular sector from $O$ to $P$.

Similarly, if $t$ is a real number, the point $P(\cosh t,\sinh t)$ lies on the right branch of the unit hyperbola $x^2-y^2=1$ (since ${\cosh }^{2}t-{\sinh }^{2}t=1$ and $\cosh t\ge 1$). The number $t$ equals twice the area of the corresponding hyperbolic sector.

Derivatives of Hyperbolic Functions

\boxed{\begin{aligned} \frac{d}{dx}\sinh x&=\cosh x\\[4pt] \frac{d}{dx}\cosh x&=\sinh x\\[4pt] \frac{d}{dx}\tanh x&=\text{sech}^2 x=1-\tanh^2 x \end{aligned}}\boxed{\begin{aligned} \frac{d}{dx}\coth x&=-\text{csch}^2 x\\[4pt] \frac{d}{dx}\text{sech}\,x&=-\text{sech}\,x\tanh x\\[4pt] \frac{d}{dx}\text{csch}\,x&=-\text{csch}\,x\coth x \end{aligned}}

Inverse Hyperbolic Functions

Since $\sinh$ and $\tanh$ are one-to-one, they have inverses. $\cosh$ is one-to-one on $[0,+\mathrm{\infty })$, where we define its inverse.

The inverses are denoted $\text{arcsinh}$, $\text{arccosh}$, $\text{arctanh}$ (or ${\sinh }^{-1}$, ${\cosh }^{-1}$, ${\tanh }^{-1}$):

\boxed{\begin{aligned} y=\sinh x&\Leftrightarrow x=\text{arcsinh}\,y &&(x,y\in\mathbb{R})\\ y=\cosh x&\Leftrightarrow x=\text{arccosh}\,y &&(x\geq0,y\geq1)\\ y=\tanh x&\Leftrightarrow x=\text{arctanh}\,y &&(x\in\mathbb{R},-1

Explicit formulas:

\boxed{\begin{aligned} \text{arcsinh}\,x&=\ln(x+\sqrt{x^2+1}) &&(x\in\mathbb{R})\\ \text{arccosh}\,x&=\ln(x+\sqrt{x^2-1}) &&(x\geq1)\\ \text{arctanh}\,x&=\frac{1}{2}\ln\frac{1+x}{1-x} &&(-1\boxed{\begin{aligned} \text{arccoth}\,x&=\frac{1}{2}\ln\frac{x+1}{x-1} &&(|x|>1)\\ \text{arcsech}\,x&=\ln\!\left(\frac{1+\sqrt{1-x^2}}{x}\right) &&(00)\\[2pt]-\ln\!\left(\frac{1+\sqrt{1+x^2}}{-x}\right) & (x<0)\end{cases} \end{aligned}}

Show $\text{arcsinh}x=\ln (x+\sqrt{x^2+1})$.

Solution

Start from $y=\sinh x=(e^x-e^{-x})/2$. Multiplying by $2e^x$: $$(e^x{)}^2-2y{e}^x-1=0\Rightarrow e^x=y+\sqrt{y^2+1}$$ (taking the positive root since ). Taking $\ln$: $$x=\ln (y+\sqrt{y^2+1})$$ Replacing $y$ by $x$: ${\sinh }^{-1}(x)=\ln (x+\sqrt{x^2+1})$.

Derivatives of Inverse Hyperbolic Functions

\boxed{\begin{aligned} \frac{d}{dx}\text{arcsinh}\,x&=\frac{1}{\sqrt{x^2+1}} &&(x\in\mathbb{R})\\[4pt] \frac{d}{dx}\text{arccosh}\,x&=\frac{1}{\sqrt{x^2-1}} &&(x>1)\\[4pt] \frac{d}{dx}\text{arctanh}\,x&=\frac{1}{1-x^2} &&(-1\boxed{\begin{aligned} \frac{d}{dx}\text{arccoth}\,x&=\frac{1}{1-x^2} &&(|x|>1)\\[4pt] \frac{d}{dx}\text{arcsech}\,x&=\frac{-1}{x\sqrt{1-x^2}} &&(00)\\[4pt]\dfrac{1}{x\sqrt{1+x^2}} & (x<0)\end{cases} \end{aligned}}

Show ${\displaystyle \frac{d}{dx}}\text{arcsinh}x={\displaystyle \frac{1}{\sqrt{x^2+1}}}$.

Solution

Let $y=\sinh x$, so $x=\text{arcsinh}y$: $$\frac{d}{dy}\text{arcsinh}y=\frac{1}{\cosh x}=\frac{1}{\sqrt{{\sinh }^{2}x+1}}=\frac{1}{\sqrt{y^2+1}}$$ (using ${\cosh }^{2}x-{\sinh }^{2}x=1$ and ) Replacing $y$ by $x$: ${\displaystyle \frac{d}{dx}}\text{arcsinh}x={\displaystyle \frac{1}{\sqrt{x^2+1}}}$.

Show ${\displaystyle \frac{d}{dx}}\text{arccosh}x={\displaystyle \frac{1}{\sqrt{x^2-1}}}$ for .

Solution

Let $y=\cosh x$ (), so $x=\text{arccosh}y$: $$\frac{d}{dy}(\text{arccosh}y)=\frac{1}{\sinh x}=\frac{1}{\sqrt{{\cosh }^{2}x-1}}=\frac{1}{\sqrt{y^2-1}}$$

Show ${\displaystyle \frac{d}{dx}}\text{arctanh}x={\displaystyle \frac{1}{1-x^2}}$ for .

Solution

Let $y=\tanh x$, so $x=\text{arctanh}y$: $$\frac{d}{dy}(\text{arctanh}y)=\frac{1}{{\text{sech}}^{2}x}=\frac{1}{1-{\tanh }^{2}x}=\frac{1}{1-y^2}$$ Replacing $y$ by $x$: ${\displaystyle \frac{d}{dx}}\text{arctanh}x={\displaystyle \frac{1}{1-x^2}}$.

Frequently Asked Questions

How are hyperbolic functions related to a hyperbola?

The point $(\cosh t,\sinh t)$ always lies on the unit hyperbola $x^2-y^2=1$, just as $(\cos \theta ,\sin \theta )$ lies on the unit circle. The parameter $t$ equals twice the area of the hyperbolic sector, analogous to $\theta$ being twice the area of the circular sector.

Do hyperbolic functions satisfy similar identities to trigonometric ones?

Yes, with one sign change: while ${\cos }^{2}x+{\sin }^{2}x=1$, we have ${\cosh }^{2}x-{\sinh }^{2}x=1$. The addition formulas also hold with the same signs. This is called the Osborn rule.

Where do hyperbolic functions appear in applications?

The shape of a hanging cable (catenary) is described by $y=a\cosh (x/a)$. Hyperbolic functions also arise in the solution of differential equations in physics and engineering, and in the theory of special relativity.