2.9 Calculus of the Hyperbolic Functions

Big idea

Any function can be split into an even part and an odd part by averaging it with its own reflection. Applied to $e^x$, that split produces two functions: the even part $\tfrac{1}{2}(e^x + e^{-x})$ and the odd part $\tfrac{1}{2}(e^x - e^{-x})$. These are the hyperbolic cosine and hyperbolic sine. They are not new functions in any deep sense, since everything about them can be unpacked into exponentials, but they combine so often that they are worth naming and worth having derivative rules for.

The names come from a resemblance rather than a coincidence. Ordinary cosine and sine satisfy $\cos^2 t + \sin^2 t = 1$, so the point $(\cos t, \sin t)$ traces the unit circle. The hyperbolic pair satisfies $\cosh^2 t - \sinh^2 t = 1$, so the point $(\cosh t, \sinh t)$ traces one branch of the hyperbola $x^2 - y^2 = 1$. That single sign change propagates through the whole subject: most identities look like the trigonometric ones with a sign flipped somewhere, and most derivative formulas look like the trigonometric ones with a minus sign missing.

They also show up on their own terms. A flexible chain hanging under its own weight settles into the shape of a hyperbolic cosine, called a catenary, not a parabola. Arc length integrals simplify dramatically for that curve because $1 + \sinh^2 x = \cosh^2 x$ removes the radical exactly, in the same way $1 + \tan^2 x = \sec^2 x$ does for a different family. The inverse hyperbolic functions supply antiderivatives for $\tfrac{1}{\sqrt{x^2+1}}$ and $\tfrac{1}{1-x^2}$, filling in gaps in the table the same way the natural logarithm filled the gap at $\tfrac{1}{x}$.

The practical approach is simple. Learn the two definitions, the one identity, and the two derivatives, and derive anything else you need on the spot. Since the definitions are just exponentials, every claim in this section is checkable by substitution, which is a luxury the trigonometric functions do not offer.

Decoder

The hyperbolic cosine and sine are the even and odd parts of the exponential function, they satisfy the identity that the square of the cosine minus the square of the sine is one, and each is the derivative of the other.

That last clause is the whole of the differential calculus here, and it is where the sign trap lives. For the circular functions, differentiating cosine produces a minus sign. For the hyperbolic functions it does not:

$$ \frac{d}{dx}\sinh x = \cosh x, \qquad \frac{d}{dx}\cosh x = \sinh x $$

Both follow in one line from the definitions, since differentiating $e^{-x}$ flips its sign and swaps the plus and minus in the two formulas. If you ever doubt a hyperbolic identity, write it in exponentials and check; it takes ten seconds and it is decisive.

The identity itself deserves the same treatment. It is not asserted, it is computed: $(e^x + e^{-x})^2 - (e^x - e^{-x})^2 = 4 e^x e^{-x} = 4$, and dividing by the $4$ in the denominators leaves $1$. Notice it is a difference, not a sum, so unlike $\sin$ and $\cos$ the hyperbolic functions are not bounded. Values of $\cosh$ start at $1$ and grow without bound, and $\sinh$ runs over all real numbers.

Definitions and results

The two basic functions. For all real $x$,

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

So $\cosh$ is even and $\sinh$ is odd, $\cosh 0 = 1$, $\sinh 0 = 0$, and $\cosh x \ge 1$ for every $x$.

The rest of the family. Defined exactly as for the circular functions:

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

Since $\cosh$ is never zero, $\tanh$ and $\operatorname{sech}$ are defined for all real numbers, and $|\tanh x| < 1$ always.

The fundamental identity. For all real $x$,

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

Dividing through by $\cosh^2 x$ gives $1 - \tanh^2 x = \operatorname{sech}^2 x$.

Sums and doubles. From the definitions,

$$ \sinh(2x) = 2\sinh x \cosh x, \qquad \cosh(2x) = \cosh^2 x + \sinh^2 x = 1 + 2\sinh^2 x $$

The last form rearranges to $\sinh^2 x = \tfrac{1}{2}\big(\cosh(2x) - 1\big)$, which is how you integrate a square.

Derivatives. With no minus signs on the first two:

$$ \frac{d}{dx}\sinh x = \cosh x, \qquad \frac{d}{dx}\cosh x = \sinh x, \qquad \frac{d}{dx}\tanh x = \operatorname{sech}^2 x $$

and $\dfrac{d}{dx}\operatorname{sech} x = -\operatorname{sech} x \tanh x$. Each combines with the chain, product and quotient rules in the usual way.

Basic integrals. Reading the derivatives backwards,

$$ \int \sinh x\, dx = \cosh x + C, \qquad \int \cosh x\, dx = \sinh x + C, \qquad \int \operatorname{sech}^2 x\, dx = \tanh x + C $$

Also $\displaystyle\int \tanh x\, dx = \ln(\cosh x) + C$, by substituting $u = \cosh x$; no absolute value is needed because $\cosh$ is positive.

Inverse functions. Since $\sinh$ is strictly increasing on the whole line it has an inverse everywhere, and solving the defining quadratic in $e^x$ gives the closed form

$$ \sinh^{-1} x = \ln\!\left( x + \sqrt{x^2 + 1} \right) $$

Differentiating the inverses supplies two antiderivatives:

$$ \frac{d}{dx}\sinh^{-1} x = \frac{1}{\sqrt{x^2+1}}, \qquad \frac{d}{dx}\tanh^{-1} x = \frac{1}{1 - x^2} \quad (|x| < 1) $$

The catenary. A uniform flexible cable hanging between two supports takes the shape $y = a \cosh(x/a)$. Because $1 + \sinh^2 = \cosh^2$, arc length along it can be computed in closed form.

Worked examples

Checking the identity and a value

Verify $\cosh^2 x - \sinh^2 x = 1$ from the definitions, then evaluate both functions at $x = \ln 2$.

Expand the squares over the common denominator $4$:

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

The numerator is a difference of squares in disguise. Using $(a+b)^2 - (a-b)^2 = 4ab$ with $a = e^x$ and $b = e^{-x}$, the numerator is $4e^x e^{-x} = 4$, so the whole expression is $1$.

At $x = \ln 2$ the exponentials are $e^{\ln 2} = 2$ and $e^{-\ln 2} = \tfrac{1}{2}$, so

$$ \cosh(\ln 2) = \frac{2 + \tfrac{1}{2}}{2} = \frac{5}{4}, \qquad \sinh(\ln 2) = \frac{2 - \tfrac{1}{2}}{2} = \frac{3}{4} $$

The identity checks numerically: $\left(\tfrac{5}{4}\right)^2 - \left(\tfrac{3}{4}\right)^2 = \tfrac{25 - 9}{16} = 1$.

A product with a chain inside

Differentiate $y = e^{2x}\cosh(3x)$.

This is a product whose second factor is a composition, so the product rule governs the outside and the chain rule handles $\cosh(3x)$, whose derivative is $3\sinh(3x)$:

$$ \frac{dy}{dx} = 2e^{2x}\cosh(3x) + e^{2x} \cdot 3\sinh(3x) = e^{2x}\big( 2\cosh(3x) + 3\sinh(3x) \big) $$

Check at $x = 0$, where $\cosh 0 = 1$ and $\sinh 0 = 0$: the formula gives $1 \cdot (2 + 0) = 2$. That is right, because near the origin $\cosh(3x)$ is flat to first order, so the growth of $y$ comes entirely from $e^{2x}$, whose derivative at $0$ is $2$.

For contrast, differentiate $y = \tanh(x^2)$: the outer derivative is $\operatorname{sech}^2$ evaluated at $x^2$, and the inner derivative is $2x$, giving $2x\operatorname{sech}^2(x^2)$.

A definite integral

Evaluate $\displaystyle\int_0^{\ln 2} \sinh x\, dx$ and $\displaystyle\int_0^{\ln 2}\cosh x\, dx$.

The antiderivatives swap the two functions, with no sign changes:

$$ \int_0^{\ln 2} \sinh x\, dx = \big[ \cosh x \big]_0^{\ln 2} = \frac{5}{4} - 1 = \frac{1}{4} $$

$$ \int_0^{\ln 2} \cosh x\, dx = \big[ \sinh x \big]_0^{\ln 2} = \frac{3}{4} - 0 = \frac{3}{4} $$

Both answers pass a size check. On $[0, \ln 2]$, an interval of length about $0.693$, the function $\sinh$ climbs from $0$ to $0.75$, so an average value around $0.36$ and an integral around $0.25$ are right. Over the same interval $\cosh$ stays between $1$ and $1.25$, so its integral should be a little above $0.693$, and $0.75$ is.

Length of a hanging cable

Find the length of the catenary $y = \cosh x$ from $x = 0$ to $x = 1$.

The arc length integrand is $\sqrt{1 + [y']^2}$, and here $y' = \sinh x$, so the identity applies directly:

$$ 1 + \sinh^2 x = \cosh^2 x $$

Since $\cosh x$ is positive, its square root is itself, and the radical disappears with no approximation:

$$ L = \int_0^1 \cosh x\, dx = \big[ \sinh x \big]_0^1 = \sinh 1 = \frac{e - e^{-1}}{2} \approx 1.175 $$

Check against the straight line. The endpoints are $(0, 1)$ and $(1, \cosh 1) \approx (1, 1.543)$, and the chord between them has length $\sqrt{1 + (0.543)^2} \approx 1.138$. The curve is longer than the chord by a small margin, which is what a gently sagging cable should give.

Practice

First the derivatives. These mix the hyperbolic functions with the product, quotient and chain rules, so decide which rule governs the outermost structure before differentiating anything, and remember that differentiating hyperbolic cosine produces no minus sign.

Then the antiderivatives, alongside the exponential and logarithmic ones they are built from. Watch for integrands that are a derivative of an inner function times something, since substitution handles most of these in one step.

Quiz

Five items on the exponential definitions, the fundamental identity, derivatives of hyperbolic expressions, and the standard integrals.