Hyperbolic Functions: sinh, cosh, tanh Explained

What Are Hyperbolic Functions?

Hyperbolic Functions are a family of functions built from the exponential $e^x$, defined so that they relate to the hyperbola $x^2 - y^2 = 1$ in the same way the ordinary trigonometric functions relate to the circle $x^2 + y^2 = 1$. The three core functions are the hyperbolic sine, the hyperbolic cosine, and the hyperbolic tangent, written $\sinh$, $\cosh$, and $\tanh$ and pronounced "sinch," "cosh," and "tanch."

Their definitions are exact and short:

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

The three reciprocals complete the set. Each is defined wherever its denominator is non-zero:

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

Because $\cosh x \geq 1$ for every real $x$, the functions $\tanh x$ and $\operatorname{sech} x$ are defined for all real numbers, while $\coth x$ and $\operatorname{csch} x$ are undefined only at $x = 0$, where $\sinh x = 0$.

How Are Hyperbolic Functions Defined From e To The x?

To evaluate a hyperbolic function, substitute the value into its exponential formula. The two building blocks $e^x$ and $e^{-x}$ do all the work: adding them and halving gives $\cosh x$, subtracting them and halving gives $\sinh x$.

Example 1: Evaluate $\sinh(\ln 2)$ and $\cosh(\ln 2)$.

Use $e^{\ln 2} = 2$ and $e^{-\ln 2} = \tfrac{1}{2}$:

$$\sinh(\ln 2) = \frac{2 - \tfrac{1}{2}}{2} = \frac{\tfrac{3}{2}}{2} = \frac{3}{4} = 0.7500$$

$$\cosh(\ln 2) = \frac{2 + \tfrac{1}{2}}{2} = \frac{\tfrac{5}{2}}{2} = \frac{5}{4} = 1.2500$$

Final answer: $\sinh(\ln 2) = 0.7500$ and $\cosh(\ln 2) = 1.2500$. As a check, $\cosh^2 - \sinh^2 = \tfrac{25}{16} - \tfrac{9}{16} = \tfrac{16}{16} = 1$, exactly the identity proved in the next section.

Two symmetry facts drop straight out of the definitions:

• $\sinh(-x) = -\sinh x$, so $\sinh$ is an odd function, and its graph passes through the origin.

• $\cosh(-x) = \cosh x$, so $\cosh$ is an even function, symmetric about the vertical axis, with a lowest point $\cosh 0 = 1$.

Why Does cosh Squared Minus sinh Squared Equal 1?

The single most important identity is $\cosh^2 x - \sinh^2 x = 1$. Every hyperbolic identity descends from it, and it is the reason the word "hyperbolic" appears in the name. Here is the full derivation from the definitions, which the ranking pages state but never show.

Square each definition:

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

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

Subtract the second from the first. The $e^{2x}$ and $e^{-2x}$ terms cancel, leaving only the constants:

$$\cosh^2 x - \sinh^2 x = \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} = \frac{4}{4} = 1$$

The geometric meaning. Set $x_{\text{coord}} = \cosh t$ and $y_{\text{coord}} = \sinh t$. Then $x_{\text{coord}}^2 - y_{\text{coord}}^2 = \cosh^2 t - \sinh^2 t = 1$, so the point $(\cosh t, \sinh t)$ always lands on the hyperbola $x^2 - y^2 = 1$. As $t$ runs over all real numbers, the point traces the right branch of that curve, because $\cosh t \geq 1$ is never negative.

This is the exact hyperbolic mirror of the circle: $(\cos t, \sin t)$ satisfies $\cos^2 t + \sin^2 t = 1$ and rides the unit circle. The only change is a single sign, plus becoming minus.

How Do You Differentiate Hyperbolic Functions?

Differentiating $\sinh$ and $\cosh$ takes one line each, because $\frac{d}{dx}[e^x] = e^x$ and $\frac{d}{dx}[e^{-x}] = -e^{-x}$:

$$\frac{d}{dx}[\sinh x] = \frac{d}{dx}\left[\frac{e^x - e^{-x}}{2}\right] = \frac{e^x + e^{-x}}{2} = \cosh x$$

$$\frac{d}{dx}[\cosh x] = \frac{d}{dx}\left[\frac{e^x + e^{-x}}{2}\right] = \frac{e^x - e^{-x}}{2} = \sinh x$$

Notice what did not happen: no minus sign appeared. This is the sharpest break from circular trig, where $\frac{d}{dx}[\cos x] = -\sin x$. For hyperbolic cosine there is no sign flip.

The hyperbolic tangent derivative is worth deriving in full, since the result $\operatorname{sech}^2 x$ is easy to misremember. Apply the quotient rule to $\tanh x = \frac{\sinh x}{\cosh x}$ and use the two derivatives above:

$$\frac{d}{dx}[\tanh x] = \frac{\cosh x \cdot \cosh x - \sinh x \cdot \sinh x}{\cosh^2 x} = \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} = \frac{1}{\cosh^2 x} = \operatorname{sech}^2 x$$

The numerator collapses to $1$ by the fundamental identity, which is why the answer is so clean.

The geometric interpretation. Each derivative is still a slope of a tangent line. At $x = 0$, for instance, $\frac{d}{dx}[\sinh x] = \cosh 0 = 1$, so the graph of $\sinh x$ crosses the origin at slope $1$, matching the slope of $y = x$ there. And because $\frac{d}{dx}[\tanh x] = \operatorname{sech}^2 x$ is always positive but shrinks toward $0$ for large $x$, the $\tanh$ curve rises everywhere yet flattens toward its horizontal asymptotes at $y = \pm 1$.

The full first-derivative set, all consistent with the derivative rules of exponentials:

Table: Derivatives of the six hyperbolic functions.

Reading the table backwards gives the matching integrals: $\int \cosh x , dx = \sinh x + C$ and $\int \sinh x , dx = \cosh x + C$, each with its constant of integration.

Example 2: Differentiate $y = \sinh(3x)$ and integrate $\int \cosh(2x), dx$.

For the derivative, apply the chain rule, the inner function is $3x$ with inner derivative $3$:

$$\frac{d}{dx}[\sinh(3x)] = \cosh(3x) \cdot 3 = 3\cosh(3x)$$

For the integral, reverse the chain rule and divide by the inner derivative $2$:

$$\int \cosh(2x), dx = \frac{1}{2}\sinh(2x) + C$$

Verify by differentiating the antiderivative back: $\frac{d}{dx}\left[\tfrac{1}{2}\sinh(2x) + C\right] = \tfrac{1}{2}\cdot 2\cosh(2x) = \cosh(2x)$, which is the original integrand.

Final answer: $\frac{d}{dx}[\sinh(3x)] = 3\cosh(3x)$ and $\int \cosh(2x), dx = \frac{1}{2}\sinh(2x) + C$.

What Are The Key Hyperbolic Identities?

Because every hyperbolic function is an exponential in disguise, the identities look like the trigonometric ones with a handful of sign changes. The table pairs each hyperbolic identity with its circular cousin so the differences are visible at a glance.

Table: Hyperbolic identities and their circular counterparts.

The identity $1 - \tanh^2 x = \operatorname{sech}^2 x$ follows from the fundamental one by dividing $\cosh^2 x - \sinh^2 x = 1$ through by $\cosh^2 x$. The pattern to remember: wherever a product of two sines appears, the hyperbolic version flips the sign relative to the circular version. That single rule reproduces the whole table.

What Are Inverse Hyperbolic Functions?

Each hyperbolic function has an inverse that undoes it, and because the originals are made of exponentials, the inverses are made of logarithms. Solving $y = \sinh x$ for $x$ leads to a quadratic in $e^x$, and the positive root gives a closed form:

$$\sinh^{-1} x = \ln\left(x + \sqrt{x^2 + 1}\right), \quad \text{valid for all real } x$$

$$\cosh^{-1} x = \ln\left(x + \sqrt{x^2 - 1}\right), \quad x \geq 1$$

$$\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right), \quad |x| < 1$$

The restriction on $\cosh^{-1} x$ (only $x \geq 1$) exists because $\cosh$ never outputs a value below $1$, and the restriction $|x| < 1$ on $\tanh^{-1} x$ exists because $\tanh$ only ever outputs values strictly between $-1$ and $1$. These are ordinary inverse functions, restricted to the ranges the originals actually reach.

Their derivatives are the reason hyperbolic functions matter in integration: for example, $\frac{d}{dx}[\sinh^{-1} x] = \frac{1}{\sqrt{x^2 + 1}}$, so $\int \frac{dx}{\sqrt{x^2 + 1}} = \sinh^{-1} x + C$, an antiderivative that would otherwise need an awkward substitution.

Why Do Hyperbolic Functions Work?

Hyperbolic functions are not a second, unrelated copy of trigonometry. They are what the same algebra produces when the underlying curve is a hyperbola instead of a circle.

• They come from splitting $e^x$. Any function can be broken into an even part and an odd part. For $e^x$, the even part is $\cosh x$ and the odd part is $\sinh x$, and adding them recovers $\cosh x + \sinh x = e^x$. The functions are simply the two halves of the exponential.

• They ride the hyperbola. The point $(\cosh t, \sinh t)$ satisfies $x^2 - y^2 = 1$, so it lives on the conic section called a hyperbola, the same way $(\cos t, \sin t)$ lives on the circle.

• The parameter is an area, not an angle. For the circle, the parameter $t$ equals twice the area of the swept sector. For the hyperbola, $t$ again equals twice the area of the hyperbolic sector. This is the deep reason the two families mirror each other, and why $t$ is not measured in degrees.

• The sign change is geometry, not accident. A circle has $x^2 + y^2 = 1$ with a plus; a hyperbola has $x^2 - y^2 = 1$ with a minus. Every "flipped sign" in the identity table traces back to that one difference in the defining curve.

Seen this way, sine and cosine and $\sinh$ and $\cosh$ are two readings of one idea. The trigonometric functions measure circular motion; the hyperbolic functions measure motion along a hyperbola.

Who Discovered Hyperbolic Functions?

For a long time the hyperbola and the circle were treated as distant relatives. The insight that they carry a matching pair of functions came in the mid-1700s, from two mathematicians working just after Euler had shown how deeply the exponential runs through all of analysis.

Two figures shaped the subject into its modern form:

• Vincenzo Riccati (1707–1775, Italy) introduced the hyperbolic sine and cosine as geometric quantities and established their basic identity.

• Johann Heinrich Lambert (1728–1777, Switzerland and Germany) developed the notation and the connection to the exponential, and popularized the functions so widely that they entered the standard toolkit of analysis.

Where Are Hyperbolic Functions Used In The Real World?

The clean derivatives and the exponential core make hyperbolic functions appear wherever growth, hanging, or steady-state behavior is modeled.

• Engineering and architecture: a chain, rope, or power line hanging under its own weight forms a catenary, $y = a\cosh(x/a)$. Suspension-bridge cables, overhead transmission lines, and the shape of the Gateway Arch in St. Louis are all designed around this curve.

• Physics: the velocity of an object falling with air resistance approaches its terminal speed along a $\tanh$ curve, and the shape of a relativistic "rapidity" adds using $\tanh$ rather than ordinary addition.

• Machine learning: $\tanh x$ is a classic activation function inside neural networks, chosen because it squashes any input smoothly into the range $(-1, 1)$.

• Signal processing and heat flow: solutions to many differential equations that describe cooling, cable signals, and vibrations are written with $\cosh$ and $\sinh$.

• Special relativity: a boost between reference frames is a "hyperbolic rotation," using $\cosh$ and $\sinh$ exactly where a spatial rotation would use $\cos$ and $\sin$.

One curve, the humble hanging chain, connects a child's swing set to a suspension bridge to the mathematics of moving reference frames. The hyperbola is quietly holding all of them up.

What Are The Most Common Mistakes With Hyperbolic Functions?

These four errors account for most lost marks on hyperbolic functions, verified against the mathcentre workbook, the Cuyamaca College M280 handout, and the "derivative of tanh" and "no sign flip" points surfaced in real search queries.

Using the wrong sign in the fundamental identity.

Where it slips in:

A student writes $\sinh^2 x + \cosh^2 x = 1$ by copying the circular identity $\sin^2 x + \cos^2 x = 1$ out of habit.

Don't do this:

Do not carry the plus sign over from circular trig. That equation is false, since $\cosh x \geq 1$ already makes $\cosh^2 x \geq 1$ on its own.

The correct way:

Use a minus and put $\cosh$ first: $\cosh^2 x - \sinh^2 x = 1$. The minus is the fingerprint of the hyperbola $x^2 - y^2 = 1$.

Adding a minus sign to the derivative of cosh.

Where it slips in:

A student writes $\frac{d}{dx}[\cosh x] = -\sinh x$, reasoning by analogy with $\frac{d}{dx}[\cos x] = -\sin x$.

Don't do this:

Do not import the sign flip from circular trig. There is no negative sign in the hyperbolic cosine derivative.

The correct way:

Differentiate from the definition: $\frac{d}{dx}[\cosh x] = \frac{e^x - e^{-x}}{2} = \sinh x$, with no minus. Both $\sinh$ and $\cosh$ differentiate to the other with a plus.

Forgetting the chain-rule inner derivative (or the $+C$).

Where it slips in:

A student writes $\frac{d}{dx}[\sinh(3x)] = \cosh(3x)$, dropping the factor of $3$, or writes $\int \sinh x , dx = \cosh x$ with no constant.

Don't do this:

Do not treat the inside as if it were a bare $x$, and do not omit the constant of integration on an indefinite integral.

The correct way:

Multiply by the inner derivative: $\frac{d}{dx}[\sinh(3x)] = 3\cosh(3x)$. And close every indefinite integral with $+C$: $\int \sinh x , dx = \cosh x + C$.

Confusing hyperbolic functions with circular ones.

Where it slips in:

Because the names $\sinh$ and $\sin$ look almost identical, a student reads $\cosh$ off a unit circle or expects $\tanh x$ to be periodic.

Don't do this:

Do not treat $\sinh$, $\cosh$, $\tanh$ as periodic wave functions. They do not repeat, and $\cosh x$ never dips below $1$.

The correct way:

Anchor them to the hyperbola, not the circle: $\cosh$ grows without bound, $\sinh$ passes once through the origin, and $\tanh$ levels off toward $\pm 1$. None of them cycles.

Practice Problems On Hyperbolic Functions

Work each one, then check against the verified answer.

1. Simplify $\cosh^2(5) - \sinh^2(5)$.
   (Answer: $1$, by the fundamental identity.)

2. Find $\frac{d}{dx}[\cosh(4x)]$.
   (Answer: $4\sinh(4x)$, by the chain rule.)

3. Evaluate $\sinh 0$ and $\cosh 0$.
   (Answer: $\sinh 0 = \frac{1-1}{2} = 0$ and $\cosh 0 = \frac{1+1}{2} = 1$.)

4. Compute $\int \sinh x , dx$.
   (Answer: $\cosh x + C$; differentiate back to recover $\sinh x$.)

5. Find $\frac{d}{dx}[\tanh(x^2)]$.
   (Answer: $2x,\operatorname{sech}^2(x^2)$, chain rule with $\frac{d}{dx}[\tanh u] = \operatorname{sech}^2 u$.)

6. Evaluate $\tanh^{-1}(0)$.
   (Answer: $\frac{1}{2}\ln\frac{1+0}{1-0} = \frac{1}{2}\ln 1 = 0$.)

Where Should You Go Next After Hyperbolic Functions?

Hyperbolic functions sit at the crossroads of exponentials, geometry, and calculus, and several natural doors open from here.

1. The hyperbola. Meet the curve $x^2 - y^2 = 1$ that gives these functions their name, and see why its two branches and asymptotes shape the graphs of $\cosh$ and $\sinh$.

2. Trigonometric functions. Revisit the circular originals side by side with their hyperbolic twins, so the matching identities and the sign differences click into place.

3. The derivative. Firm up the differentiation rules that make $\frac{d}{dx}[\sinh x] = \cosh x$ and $\frac{d}{dx}[\tanh x] = \operatorname{sech}^2 x$ feel automatic.

If your child is building toward calculus, a live Bhanzu trainer teaches hyperbolic functions starting from the "why" (the hyperbola and the split of $e^x$) rather than a memorized table, in the Bhanzu math program.