Hyperbolic Functions and Their Inverses

Just as the pair $(\cos\theta, \sin\theta)$ traces the unit circle $x^2 + y^2 = 1$ as $\theta$ varies, the hyperbolic functions parametrize the unit hyperbola $x^2 - y^2 = 1$: the point $(\cosh t, \sinh t)$ lies on that hyperbola for every real $t$. Beyond the geometric picture, they arise naturally as solutions to the differential equation $y'' = y$, describe the shape of a hanging cable (the catenary), and appear in special relativity through Lorentz boosts. Unlike their trigonometric counterparts, they are not periodic — they are built directly from the exponential function.

Definitions

Hyperbolic cosine and hyperbolic sine are defined by

$$\cosh x \coloneqq \frac{e^x + e^{-x}}{2}, \qquad \sinh x \coloneqq \frac{e^x - e^{-x}}{2}. \tag{1}$$

You can think of $\cosh x$ as the even part and $\sinh x$ as the odd part of the exponential, since $e^x = \cosh x + \sinh x$ and $e^{-x} = \cosh x - \sinh x$.

The remaining four hyperbolic functions are defined as ratios built from these two:

$$\tanh x \coloneqq \frac{\sinh x}{\cosh x}, \qquad \operatorname{coth} x \coloneqq \frac{\cosh x}{\sinh x},$$ $$\operatorname{sech} x \coloneqq \frac{1}{\cosh x}, \qquad \operatorname{csch} x \coloneqq \frac{1}{\sinh x}.$$

Note that $\operatorname{coth} x$ and $\operatorname{csch} x$ are undefined at $x = 0$, where $\sinh 0 = 0$.

Fundamental identity

The central algebraic relation among the hyperbolic functions is

$$\cosh^2 x - \sinh^2 x = 1.$$

Derivation. Substituting from definitions (1):

$$\cosh^2 x - \sinh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^{\!2} - \left(\frac{e^x - e^{-x}}{2}\right)^{\!2} = \frac{(e^x + e^{-x})^2 - (e^x - e^{-x})^2}{4}.$$

Using the algebraic identity $(a + b)^2 - (a - b)^2 = 4ab$ with $a = e^x$ and $b = e^{-x}$, the numerator becomes $4e^x e^{-x} = 4e^0 = 4$. Dividing by $4$ gives $1$. $\square$

This mirrors the circular identity $\cos^2\theta + \sin^2\theta = 1$, and it confirms that $(\cosh t,\, \sinh t)$ satisfies $x^2 - y^2 = 1$ for every real $t$.

Addition formulas

For all real $x$ and $y$:

$$\cosh(x + y) = \cosh x \cosh y + \sinh x \sinh y,$$ $$\sinh(x + y) = \sinh x \cosh y + \cosh x \sinh y.$$

You can verify these by substituting definitions (1) and expanding. Compare with the circular formulas $\cos(x+y) = \cos x\cos y - \sin x\sin y$ and $\sin(x+y) = \sin x\cos y + \cos x\sin y$: the $\cosh$ addition formula has a $+$ sign where the cosine formula has a $-$. This sign flip is a direct consequence of the sign difference in the fundamental identity ($\cosh^2 - \sinh^2 = 1$ versus $\cos^2 + \sin^2 = 1$).

Derivatives

Differentiating definitions (1) term by term with respect to $x$:

$$(\cosh x)' = \frac{e^x - e^{-x}}{2} = \sinh x, \qquad (\sinh x)' = \frac{e^x + e^{-x}}{2} = \cosh x.$$

So $\sinh$ and $\cosh$ are each other’s derivatives — they swap, in contrast to $\sin$ and $\cos$, which alternate with a sign change under each differentiation.

For $\tanh$, apply the quotient rule and then use the fundamental identity:

$$(\tanh x)' = \frac{(\sinh x)'\cosh x - \sinh x\,(\cosh x)'}{\cosh^2 x} = \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} = \frac{1}{\cosh^2 x} = \operatorname{sech}^2 x.$$

Applying the fundamental identity once more gives the alternative form $\operatorname{sech}^2 x = 1 - \tanh^2 x$, so

$$(\tanh x)' = \operatorname{sech}^2 x = 1 - \tanh^2 x.$$

The derivatives of the remaining three functions follow from the quotient and chain rules:

$$(\operatorname{coth} x)' = -\operatorname{csch}^2 x, \qquad (\operatorname{sech} x)' = -\operatorname{sech} x\tanh x, \qquad (\operatorname{csch} x)' = -\operatorname{csch} x\operatorname{coth} x.$$

Properties of cosh and sinh

• $\cosh$ is even: $\cosh(-x) = \cosh x$. By the AM–GM inequality, $e^x + e^{-x} \geq 2\sqrt{e^x \cdot e^{-x}} = 2$, so $\cosh x \geq 1$ for all $x \in \mathbb{R}$, with equality only at $x = 0$. The graph of $y = \cosh x$ is a catenary — the shape assumed by a uniform flexible chain hanging under gravity.
• $\sinh$ is odd: $\sinh(-x) = -\sinh x$. It is strictly increasing on all of $\mathbb{R}$ (since $(\sinh x)' = \cosh x \geq 1 > 0$), with range $\mathbb{R}$.
• $\tanh$ is odd and strictly increasing, mapping $\mathbb{R}$ onto the open interval $(-1, 1)$. As $x \to \pm\infty$, $e^{-|x|} \to 0$ forces $\tanh x \to \pm 1$, so $y = \pm 1$ are horizontal asymptotes.

Inverse hyperbolic functions

Because $\sinh$ is strictly increasing on $\mathbb{R}$, it has a global inverse. For $\cosh$, you must restrict to $[0, \infty)$, where it is strictly increasing. For $\tanh$ the natural domain is $\mathbb{R}$ with range $(-1, 1)$.

The remarkable feature is that all three inverses have closed-form expressions in terms of the logarithm.

Arsinh

Set $y = \sinh x = \dfrac{e^x - e^{-x}}{2}$ and solve for $x$. Multiplying both sides by $2e^x$:

$$e^{2x} - 2y\,e^x - 1 = 0.$$

This is a quadratic in $e^x$. The quadratic formula gives $e^x = y \pm \sqrt{y^2 + 1}$. Since $e^x > 0$ and $\sqrt{y^2+1} > |y|$, only the positive root is admissible. Taking logarithms:

$$\operatorname{arsinh} x \coloneqq \ln\!\left(x + \sqrt{x^2 + 1}\right), \qquad x \in \mathbb{R}.$$

Arcosh

Set $y = \cosh x = \dfrac{e^x + e^{-x}}{2}$ with $x \geq 0$ and solve. Multiplying by $2e^x$:

$$e^{2x} - 2y\,e^x + 1 = 0,$$

so $e^x = y \pm \sqrt{y^2 - 1}$. This requires $y \geq 1$. For $x \geq 0$ the larger root corresponds to $x \geq 0$, so we choose the $+$ sign. Taking logarithms:

$$\operatorname{arcosh} x \coloneqq \ln\!\left(x + \sqrt{x^2 - 1}\right), \qquad x \geq 1.$$

Artanh

Set $y = \tanh x = \dfrac{e^x - e^{-x}}{e^x + e^{-x}}$ and solve. Writing $u = e^{2x}$:

$$y = \frac{u - 1}{u + 1} \quad\Longrightarrow\quad y(u + 1) = u - 1 \quad\Longrightarrow\quad u = \frac{1 + y}{1 - y}.$$

Since $u = e^{2x}$, taking logarithms gives $2x = \ln\!\dfrac{1+y}{1-y}$, so:

$$\operatorname{artanh} x \coloneqq \frac{1}{2}\ln\frac{1 + x}{1 - x}, \qquad |x| < 1.$$

The restriction $|x| < 1$ matches the range of $\tanh$ and keeps both $1 + x$ and $1 - x$ strictly positive inside the logarithm.

Derivatives of inverse hyperbolic functions

You can find these derivatives either by differentiating the logarithmic expressions directly or by applying the inverse function theorem. Both routes are shown below.

Derivative of arsinh

Differentiating $\operatorname{arsinh} x = \ln\!\left(x + \sqrt{x^2+1}\right)$:

$$(\operatorname{arsinh} x)' = \frac{1}{x + \sqrt{x^2+1}} \cdot \left(1 + \frac{x}{\sqrt{x^2+1}}\right) = \frac{1}{x + \sqrt{x^2+1}} \cdot \frac{x + \sqrt{x^2+1}}{\sqrt{x^2+1}} = \frac{1}{\sqrt{x^2+1}}.$$ $$\operatorname{arsinh}'(x) = \frac{1}{\sqrt{x^2 + 1}}, \qquad x \in \mathbb{R}.$$

Derivative of arcosh

Differentiating $\operatorname{arcosh} x = \ln\!\left(x + \sqrt{x^2-1}\right)$ for $x > 1$:

$$(\operatorname{arcosh} x)' = \frac{1}{x + \sqrt{x^2-1}} \cdot \left(1 + \frac{x}{\sqrt{x^2-1}}\right) = \frac{1}{x + \sqrt{x^2-1}} \cdot \frac{x + \sqrt{x^2-1}}{\sqrt{x^2-1}} = \frac{1}{\sqrt{x^2-1}}.$$ $$\operatorname{arcosh}'(x) = \frac{1}{\sqrt{x^2 - 1}}, \qquad x > 1.$$

Derivative of artanh

Rewrite $\operatorname{artanh} x = \dfrac{1}{2}\ln(1+x) - \dfrac{1}{2}\ln(1-x)$ and differentiate for $|x| < 1$:

$$(\operatorname{artanh} x)' = \frac{1}{2} \cdot \frac{1}{1+x} + \frac{1}{2} \cdot \frac{1}{1-x} = \frac{1}{2} \cdot \frac{(1-x) + (1+x)}{(1+x)(1-x)} = \frac{1}{1 - x^2}.$$ $$\operatorname{artanh}'(x) = \frac{1}{1 - x^2}, \qquad |x| < 1.$$

Compare this with $\arctan'(x) = \dfrac{1}{1+x^2}$ from Inverse Trigonometric Functions: the only difference is the sign in the denominator, a reflection of the $\cosh^2 - \sinh^2 = 1$ identity versus $\cos^2 + \sin^2 = 1$.

Table of derivatives

Derivatives of the six hyperbolic functions, collected for reference:

Function	Derivative
$\sinh x$	$\cosh x$
$\cosh x$	$\sinh x$
$\tanh x$	$\operatorname{sech}^2 x$
$\operatorname{coth} x$	$-\operatorname{csch}^2 x$
$\operatorname{sech} x$	$-\operatorname{sech} x\tanh x$
$\operatorname{csch} x$	$-\operatorname{csch} x\operatorname{coth} x$

Summary

• The hyperbolic functions are defined via the exponential: $\cosh x \coloneqq \dfrac{e^x+e^{-x}}{2}$, $\sinh x \coloneqq \dfrac{e^x-e^{-x}}{2}$, and $\tanh x \coloneqq \dfrac{\sinh x}{\cosh x}$.
• The fundamental identity $\cosh^2 x - \sinh^2 x = 1$ mirrors the Pythagorean identity and shows that $(\cosh t, \sinh t)$ lies on the unit hyperbola.
• Derivatives: $(\sinh x)' = \cosh x$, $(\cosh x)' = \sinh x$, $(\tanh x)' = \operatorname{sech}^2 x$.
• $\cosh$ is even and satisfies $\cosh x \geq 1$; $\sinh$ is odd and strictly increasing with range $\mathbb{R}$; $\tanh$ maps $\mathbb{R}$ onto $(-1, 1)$.
• The inverse hyperbolic functions have closed-form logarithmic expressions derived by solving for $x$ algebraically:
  • $\operatorname{arsinh} x = \ln\!\left(x + \sqrt{x^2+1}\right)$, defined on $\mathbb{R}$.
  • $\operatorname{arcosh} x = \ln\!\left(x + \sqrt{x^2-1}\right)$, defined on $[1, \infty)$.
  • $\operatorname{artanh} x = \dfrac{1}{2}\ln\dfrac{1+x}{1-x}$, defined on $(-1, 1)$.
• Their derivatives — $\dfrac{1}{\sqrt{x^2+1}}$, $\dfrac{1}{\sqrt{x^2-1}}$, $\dfrac{1}{1-x^2}$ — closely parallel those of the inverse trigonometric functions, differing only in the signs under the square root and in the denominator.