---
title: Growth, Decay, Inverse Trigonometric, and Hyperbolic Functions
module: Exponential, Logarithmic, and Inverse Functions
moduleNumber: 6
lessonNumber: 2
order: 602
summary: >
  Any quantity whose rate of change is proportional to its size grows or decays
  exponentially, the single equation y' = ky behind populations, radioactive
  decay, cooling, and continuously compounded interest. The inverse
  trigonometric functions have algebraic derivatives, and the hyperbolic
  functions, built from e^x and e^{-x}, describe the hanging cable.
topics: [Exponential, Logarithmic, and Inverse Functions]
sources:
  - book: Stewart
    ref: "Ch. 6 — Inverse Functions; §6.5 Exponential Growth and Decay"
  - book: Stewart
    ref: "§6.6 Inverse Trigonometric Functions"
  - book: Stewart
    ref: "§6.7 Hyperbolic Functions"
draft: false
---

In many natural processes a quantity changes at a rate proportional to how much
of it is present: a population reproduces in proportion to its size, a radioactive sample
decays in proportion to its mass, an investment earns interest in proportion to
its balance. Each is governed by $\d y/\d t = ky$, and the only functions solving
that equation are exponentials. Solving it accounts for radioactive dating,
cooling, and compound interest; the same
[inverse-function calculus](/calculus/exponential-logarithmic-and-inverse-functions/inverse-functions-logarithms-and-exponentials)
then produces the inverse trigonometric and hyperbolic functions and their
derivatives.

## The law of natural growth

> **Definition (Natural growth equation).** A quantity $y(t)$ obeys the **law of
> natural growth** (or decay) if its rate of change is proportional to its size:
> $$
> \frac{\d y}{\d t} = ky
> $$
> for a constant $k$. The constant is a growth rate if $k > 0$ and a decay rate
> if $k < 0$. This is a **differential equation**: it relates an unknown
> function to its derivative.

A solution is a function whose derivative is a constant multiple of itself — a
description the natural exponential fits. Any $y(t) = Ce^{kt}$
satisfies
$$
y'(t) = Cke^{kt} = k\bigl(Ce^{kt}\bigr) = ky(t).
$$
Setting $t = 0$ gives $y(0) = C$, identifying the constant as the initial value.
It can be shown that these are the only solutions.

> **Theorem (Solution of the growth equation).** The only solutions of
> $\d y/\d t = ky$ are
> $$
> y(t) = y(0)\, e^{kt}.
> $$

For a population $P(t)$ the equation is often written $\frac{1}{P}\frac{\d P}{\d t} = k$:
the growth rate divided by the population, the **relative growth rate**, is
constant. A population with constant relative growth rate grows exponentially,
and $k$ is the fractional increase per unit time. If $\d P/\d t = 0.02P$ with $t$ in
years, the population grows at a relative rate of $2\%$ per year and
$P(t) = P_0 e^{0.02t}$.

$$
% caption: Solutions of y' = ky are exponentials. A positive rate k gives
% growth; a negative rate gives decay toward zero. Both share the value y(0).
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-0.3,0) -- (4.6,0) node[right] {$t$};
\draw[->, black] (0,-0.3) -- (0,3.9) node[above] {$y$};
\draw[acc, very thick] plot[smooth, domain=0:1.75] (\x, {pow(2.71828, 0.72*\x)}) node[above left] {$k>0$};
\draw[acc, very thick, densely dashed] plot[smooth, domain=0:4.15] (\x, {pow(2.71828, -0.55*\x)}) node[above right] {$k<0$};
\fill[black] (0,1) circle (2pt) node[anchor=east] {$y(0)$};
\end{tikzpicture}
$$

### Radioactive decay and half-life

A radioactive substance loses mass at a rate proportional to the mass present,
so $\d m/\d t = km$ with $k < 0$ and $m(t) = m_0 e^{kt}$. Physicists report the rate
through the **half-life**, the time for half of any quantity to decay.

> **Worked example.** Radium-226 has a half-life of $1590$ years. Starting from
> $100$ mg, when does the sample fall to $30$ mg?
>
> Fix $k$ from the half-life. With $m_0 = 100$, the condition $m(1590) = 50$
> gives
>
> $$
> 100\, e^{1590k} = 50
> \quad\Longrightarrow\quad
> 1590k = \ln\tfrac{1}{2} = -\ln 2
> \quad\Longrightarrow\quad
> k = -\frac{\ln 2}{1590},
> $$
>
> so $m(t) = 100\, e^{-(\ln 2)t/1590} = 100\cdot 2^{-t/1590}$. Setting
> $m(t) = 30$ and solving,
>
> $$
> e^{-(\ln 2)t/1590} = 0.3
> \quad\Longrightarrow\quad
> t = -1590\,\frac{\ln 0.3}{\ln 2} \approx 2762 \text{ years}.
> $$

Radiocarbon dating runs the same computation in reverse: measure the fraction of
$^{14}\text{C}$ remaining and invert $m(t)/m_0 = e^{kt}$ to recover the age.

> **Worked example.** Carbon-14 has a half-life of $5730$ years. A fossil retains
> $30\%$ of its original $^{14}\text{C}$. How old is it?
>
> The decay constant comes from the half-life exactly as above,
> $k = -(\ln 2)/5730$, so $m(t)/m_0 = e^{kt} = 2^{-t/5730}$. The measured
> fraction is $0.30$:
>
> $$
> 2^{-t/5730} = 0.30
> \quad\Longrightarrow\quad
> t = -5730\,\frac{\ln 0.30}{\ln 2} \approx 9953 \text{ years}.
> $$
>
> The method reaches back tens of thousands of years; beyond about ten
> half-lives the surviving fraction is too small to measure reliably.

### Newton's law of cooling

An object cools at a rate proportional to the temperature gap between it and its
surroundings:
$$
\frac{\d T}{\d t} = k\bigl(T - T_s\bigr),
$$
where $T_s$ is the ambient temperature. The gap $y = T - T_s$ satisfies
$y' = ky$ because $T_s$ is constant, reducing the problem to natural decay, and
$T(t) = T_s + \bigl(T(0) - T_s\bigr)e^{kt}$.

$$
% caption: Newton cooling. The temperature decays exponentially toward the
% ambient level Ts, the horizontal asymptote, from the initial gap T(0) - Ts.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-0.3,0) -- (5.2,0) node[right] {$t$};
\draw[->, black] (0,-0.3) -- (0,3.5) node[above] {$T$};
\draw[black, dashed] (0,1.0) -- (5.0,1.0) node[right, black] {$T_s$};
\draw[acc, very thick] plot[smooth, domain=0:4.7] (\x, {1.0 + 2.0*exp(-0.62*\x)}) node[above right] {$T(t)$};
\fill[black] (0,3.0) circle (1.8pt) node[anchor=south east] {$T(0)$};
\end{tikzpicture}
$$

> **Worked example.** A soda at $72^{\circ}\text{F}$ is placed in a
> $44^{\circ}\text{F}$ refrigerator and cools to $61^{\circ}\text{F}$ after
> $30$ minutes. When does it reach $50^{\circ}\text{F}$?
>
> The gap starts at $y(0) = 72 - 44 = 28$. The reading at $30$ minutes fixes
> $k$: since $61 - 44 = 17$,
>
> $$
> 28\,e^{30k} = 17
> \quad\Longrightarrow\quad
> k = \tfrac{1}{30}\ln\tfrac{17}{28} \approx -0.01663,
> $$
>
> so $T(t) = 44 + 28\,e^{-0.01663t}$. Setting $T(t) = 50$ gives
> $28\,e^{-0.01663t} = 6$, hence
>
> $$
> t = \frac{\ln(6/28)}{-0.01663} \approx 92.6 \text{ minutes}.
> $$
>
> As $t \to \infty$ the exponential vanishes and $T \to 44$, the ambient
> temperature.

### Continuously compounded interest

An amount $A_0$ at annual rate $r$ compounded $n$ times a year grows to
$A_0(1 + r/n)^{nt}$ after $t$ years. Letting $n \to \infty$ compounds
continuously, and the [limit definition of $e$](/calculus/exponential-logarithmic-and-inverse-functions/inverse-functions-logarithms-and-exponentials)
collapses the expression:
$$
A(t) = \lim_{n \to \infty} A_0\left(1 + \frac{r}{n}\right)^{nt}
= A_0\left[\lim_{m \to \infty}\left(1 + \frac{1}{m}\right)^{m}\right]^{rt}
= A_0 e^{rt},
$$
with $m = n/r$. Differentiating recovers $\d A/\d t = rA$: continuous compounding is
exactly the statement that the balance grows at a rate proportional to itself.

> **Worked example.** Invest $\text{\textdollar}1000$ at $6\%$ compounded
> continuously. What is it worth after $3$ years, and what constant annual
> rate would match it?
>
> With $A_0 = 1000$, $r = 0.06$, $t = 3$,
>
> $$
> A(3) = 1000\,e^{(0.06)(3)} = 1000\,e^{0.18} \approx \text{\textdollar}1197.22.
> $$
>
> A single annual compounding to the same total needs $(1 + r_{\text{eff}})^3 =
> e^{0.18}$, so $r_{\text{eff}} = e^{0.06} - 1 \approx 0.0618$: the effective
> annual rate is $6.18\%$, slightly above the nominal $6\%$.

| Process | Equation | Sign of $k$ | Reported as |
| --- | --- | --- | --- |
| Population | $\d P/\d t = kP$ | $k > 0$ | relative growth rate |
| Radioactive decay | $\d m/\d t = km$ | $k < 0$ | half-life |
| Newton cooling | $\d T/\d t = k(T - T_s)$ | $k < 0$ | via gap $T - T_s$ |
| Continuous interest | $\d A/\d t = rA$ | $k = r > 0$ | annual rate |

## Inverse trigonometric functions

The trigonometric functions are periodic, so none is one-to-one and none has an
inverse as it stands. Restricting the domain to a single monotonic stretch
repairs this. The sine function on $[-\pi/2,\, \pi/2]$ increases from $-1$ to
$1$, passes the Horizontal Line Test, and so has an inverse.

> **Definition (Arcsine).** For $-1 \le x \le 1$,
> $$
> \sin^{-1} x = y \iff \sin y = x \ \text{ and } \ -\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2}.
> $$
> Thus $\arcsin x$ is the angle in $[-\pi/2,\, \pi/2]$ whose sine is $x$. It has
> domain $[-1, 1]$ and range $[-\pi/2,\, \pi/2]$.

The notation $\sin^{-1} x$ is the inverse function, not the reciprocal
$1/\sin x$. The cosine restricted to $[0, \pi]$ and the tangent restricted to
$(-\pi/2,\, \pi/2)$ yield $\cos^{-1}$ and $\tan^{-1}$ the same way.

$$
% caption: The sine restricted to the interval where it increases from -1 to 1
% is one-to-one; reflecting that arc across y = x produces the arcsine.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-2.2,0) -- (2.2,0) node[right] {$x$};
\draw[->, black] (0,-2.2) -- (0,2.2) node[above] {$y$};
\draw[black, dashed] (-1.9,-1.9) -- (1.9,1.9) node[right, black] {$y=x$};
% restricted sine on [-pi/2, pi/2] ~ [-1.571, 1.571]
\draw[acc, very thick] plot[smooth, domain=-1.571:1.571] (\x, {sin(deg(\x))});
\node[acc, anchor=north west] at (1.15,0.62) {$\sin x$};
% arcsine: reflection, domain [-1,1]
\draw[acc, very thick, densely dashed] plot[smooth, domain=-1:1] ({sin(deg(\x))}, \x);
\node[acc, anchor=south east] at (0.62,1.15) {$\arcsin x$};
\fill[black] (1.571,1) circle (1.6pt);
\fill[black] (1,1.571) circle (1.6pt);
\end{tikzpicture}
$$

### Derivatives by implicit differentiation

Each derivative follows from the [inverse-function method](/calculus/derivatives/implicit-differentiation-and-related-rates).
Let $y = \sin^{-1} x$, so $\sin y = x$ with $y \in [-\pi/2,\, \pi/2]$.
Differentiating implicitly,
$$
\cos y \,\frac{\d y}{\d x} = 1
\quad\Longrightarrow\quad
\frac{\d y}{\d x} = \frac{1}{\cos y}.
$$
Since $y$ lies in $[-\pi/2,\, \pi/2]$, $\cos y \ge 0$, so
$\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$, giving an algebraic derivative
with no trigonometry left in it:
$$
\frac{\d}{\d x}\,\sin^{-1} x = \frac{1}{\sqrt{1 - x^2}}.
$$
The tangent case is similar: with $y = \tan^{-1} x$ and $\tan y = x$,
$\sec^2 y\, (\d y/\d x) = 1$ and $\sec^2 y = 1 + \tan^2 y = 1 + x^2$, so the
derivative is $1/(1 + x^2)$, defined for all real $x$.

A right triangle makes the algebraic conversion visible. For arcsine, the angle
$y$ has opposite leg $x$ and hypotenuse $1$, so the adjacent leg is
$b = \sqrt{1 - x^2}$ and $\cos y = b = \sqrt{1 - x^2}$. For arctangent, the legs
are $1$ and $x$, so the hypotenuse is $c = \sqrt{1 + x^2}$ and
$\sec y = c = \sqrt{1 + x^2}$.

$$
% caption: Reference triangles for the inverse-trig derivatives. Left (arcsine):
% opposite leg x, hypotenuse 1, so adjacent b = root(1 - x^2) and cos y = b.
% Right (arctangent): legs 1 and x, so hypotenuse c = root(1 + x^2).
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\begin{scope}
  \draw[acc, very thick] (0,0) -- (2.4,0) -- (2.4,1.5) -- cycle;
  \draw[black] (2.2,0) -- (2.2,0.2) -- (2.4,0.2);
  \node[anchor=north] at (1.2,-0.02) {$b$};
  \node[anchor=west] at (2.46,0.75) {$x$};
  \node[anchor=south east] at (1.12,0.82) {$1$};
  \node[acc, anchor=west] at (0.3,0.16) {$y$};
\end{scope}
\begin{scope}[xshift=5.6cm]
  \draw[acc, very thick] (0,0) -- (2.4,0) -- (2.4,1.5) -- cycle;
  \draw[black] (2.2,0) -- (2.2,0.2) -- (2.4,0.2);
  \node[anchor=north] at (1.2,-0.02) {$1$};
  \node[anchor=west] at (2.46,0.75) {$x$};
  \node[anchor=south east] at (1.12,0.82) {$c$};
  \node[acc, anchor=west] at (0.3,0.16) {$y$};
\end{scope}
\end{tikzpicture}
$$

The same triangles evaluate compositions of a trig function with a different
inverse trig function.

> **Worked example.** Simplify $\cos(\tan^{-1} x)$.
>
> Let $y = \tan^{-1} x$, so $\tan y = x$ with $-\tfrac{\pi}{2} < y < \tfrac{\pi}{2}$.
> From the arctangent triangle (right, above), $\tan y = x$ means the legs are
> $1$ and $x$, so the hypotenuse is $\sqrt{1 + x^2}$ and the cosine is adjacent
> over hypotenuse:
>
> $$
> \cos(\tan^{-1} x) = \cos y = \frac{1}{\sqrt{1 + x^2}}.
> $$
>
> The same reading gives $\sin(\tan^{-1} x) = x/\sqrt{1 + x^2}$, turning a
> trigonometric composition into a plain algebraic expression.

> **Table of Derivatives.**
> $$
> \frac{\d}{\d x}\sin^{-1} x = \frac{1}{\sqrt{1 - x^2}},
> \qquad
> \frac{\d}{\d x}\cos^{-1} x = -\frac{1}{\sqrt{1 - x^2}},
> \qquad
> \frac{\d}{\d x}\tan^{-1} x = \frac{1}{1 + x^2},
> $$
> $$
> \frac{\d}{\d x}\csc^{-1} x = -\frac{1}{x\sqrt{x^2 - 1}},
> \qquad
> \frac{\d}{\d x}\sec^{-1} x = \frac{1}{x\sqrt{x^2 - 1}},
> \qquad
> \frac{\d}{\d x}\cot^{-1} x = -\frac{1}{1 + x^2}.
> $$

The arctangent flattens toward horizontal asymptotes. Reflecting the vertical
asymptotes of $\tan x$ at $x = \pm\pi/2$ into the diagonal turns them into the
horizontal lines $y = \pm\pi/2$:
$$
\lim_{x \to \infty} \tan^{-1} x = \frac{\pi}{2},
\qquad
\lim_{x \to -\infty} \tan^{-1} x = -\frac{\pi}{2}.
$$

$$
% caption: The arctangent is defined for all x and levels off toward the two
% horizontal asymptotes, the mirror images of the tangent's vertical asymptotes.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-4.4,0) -- (4.4,0) node[right] {$x$};
\draw[->, black] (0,-1.9) -- (0,1.9) node[above] {$y$};
\draw[black, dashed] (-4.3,1.571) -- (4.3,1.571);
\draw[black, dashed] (-4.3,-1.571) -- (4.3,-1.571);
\node[black, anchor=south east] at (4.3,1.571) {asymptote};
\node[black, anchor=north east] at (4.3,-1.571) {asymptote};
\draw[acc, very thick] plot[smooth, domain=-4.2:4.2, samples=60] (\x, {rad(atan(\x))}) node[below right] {$\arctan x$};
\end{tikzpicture}
$$

### Integrals from the inverse trig derivatives

Read backward, two of these formulas fill gaps in the integral table:
$$
\int \frac{1}{\sqrt{1 - x^2}}\,\d x = \sin^{-1} x + C,
\qquad
\int \frac{1}{x^2 + 1}\,\d x = \tan^{-1} x + C.
$$
Scaling by a constant $a$ generalizes the second: substituting $u = x/a$ turns
$\int \frac{\d x}{x^2 + a^2}$ into
$$
\int \frac{1}{x^2 + a^2}\,\d x = \frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right) + C.
$$
These are the antiderivatives that arise when integrating rational functions
with irreducible quadratic denominators, a technique developed in
[partial fractions](/calculus/techniques-of-integration/partial-fractions-and-integration-strategy).

## Hyperbolic functions

Certain even and odd combinations of $e^x$ and $e^{-x}$ appear so often that
they are named. They mirror the trigonometric functions in form and relate to
the hyperbola $x^2 - y^2 = 1$ the way the circular functions relate to the
circle.

> **Definition (Hyperbolic functions).**
> $$
> \sinh x = \frac{e^x - e^{-x}}{2},
> \qquad
> \cosh x = \frac{e^x + e^{-x}}{2},
> \qquad
> \tanh x = \frac{\sinh x}{\cosh x},
> $$
> with $\operatorname{csch}$, $\operatorname{sech}$, and $\coth$ the reciprocals
> of $\sinh$, $\cosh$, and $\tanh$.

Hyperbolic sine is odd with range $\mathbb{R}$; hyperbolic cosine is even with
range $[1, \infty)$; hyperbolic tangent has the horizontal asymptotes
$y = \pm 1$. They obey identities that echo the trigonometric ones, with
occasional sign changes.

> **Hyperbolic identities.**
> $$
> \cosh^2 x - \sinh^2 x = 1,
> \qquad
> 1 - \tanh^2 x = \operatorname{sech}^2 x,
> $$
> $$
> \sinh(x + y) = \sinh x \cosh y + \cosh x \sinh y,
> \qquad
> \cosh(x + y) = \cosh x \cosh y + \sinh x \sinh y.
> $$

The first identity is a direct computation and explains the name. Squaring the
definitions,
$$
\cosh^2 x - \sinh^2 x
= \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4}
= \frac{4}{4} = 1.
$$
The point $(\cosh t,\, \sinh t)$ therefore lies on the hyperbola
$x^2 - y^2 = 1$, just as $(\cos t,\, \sin t)$ lies on the circle
$x^2 + y^2 = 1$. The parameter $t$ measures twice the area of the hyperbolic
sector, not an angle.

The derivatives come straight from the definitions and $\frac{\d}{\d x}e^{\pm x} = \pm e^{\pm x}$.
For instance,
$$
\frac{\d}{\d x}\sinh x = \frac{\d}{\d x}\,\frac{e^x - e^{-x}}{2} = \frac{e^x + e^{-x}}{2} = \cosh x.
$$

> **Derivatives of hyperbolic functions.**
> $$
> \frac{\d}{\d x}\sinh x = \cosh x,
> \qquad
> \frac{\d}{\d x}\cosh x = \sinh x,
> \qquad
> \frac{\d}{\d x}\tanh x = \operatorname{sech}^2 x,
> $$
> $$
> \frac{\d}{\d x}\operatorname{csch} x = -\operatorname{csch} x \coth x,
> \quad
> \frac{\d}{\d x}\operatorname{sech} x = -\operatorname{sech} x \tanh x,
> \quad
> \frac{\d}{\d x}\coth x = -\operatorname{csch}^2 x.
> $$

Note the missing minus sign in $\frac{\d}{\d x}\cosh x = +\sinh x$, unlike the
circular case where cosine differentiates to $-\sin$.

### The catenary

The most familiar application is the shape of a hanging cable. A heavy flexible
chain suspended between two points of equal height settles into the curve
$$
y = c + a\cosh\!\left(\frac{x}{a}\right),
$$
called a **catenary**, from the Latin _catena_, chain. It resembles a parabola
but is $\cosh$: this is the profile that balances the tension along a uniformly
heavy line, and a parabola instead solves the cable of a suspension bridge,
which carries a uniform horizontal load.

$$
% caption: A hanging cable takes the catenary shape y = c + a cosh(x/a); the
% inset places the point (cosh t, sinh t) on the hyperbola x^2 - y^2 = 1.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% catenary
\draw[->, black] (-2.6,0) -- (2.7,0) node[right] {$x$};
\draw[->, black] (0,-0.3) -- (0,3.3) node[above] {$y$};
\draw[acc, very thick] plot[smooth, domain=-2.2:2.2, samples=40] (\x, {0.5*(exp(\x*0.9)+exp(-\x*0.9))}) node[above right] {$a\cosh\frac{x}{a}$};
\fill[black] (-2.2,{0.5*(exp(2.2*0.9)+exp(-2.2*0.9))}) circle (1.8pt);
\fill[black] (2.2,{0.5*(exp(2.2*0.9)+exp(-2.2*0.9))}) circle (1.8pt);
% inset hyperbola
\begin{scope}[xshift=4.9cm, yshift=1.3cm, scale=0.62]
  \draw[->, black] (-0.3,-2.0) -- (0.3,2.0);
  \draw[->, black] (-0.3,0) -- (2.6,0);
  \draw[black, thick] plot[smooth, domain=-1.3:1.3] ({0.5*(exp(\x)+exp(-\x))}, {0.5*(exp(\x)-exp(-\x))});
  \node[black, anchor=west] at (1.85,-1.55) {$x^2-y^2=1$};
  \fill[acc] ({0.5*(exp(0.75)+exp(-0.75))},{0.5*(exp(0.75)-exp(-0.75))}) circle (2.6pt);
  \node[acc, anchor=west, font=\footnotesize] at ({0.5*(exp(0.75)+exp(-0.75))+0.15},{0.5*(exp(0.75)-exp(-0.75))+0.2}) {$(\cosh t,\,\sinh t)$};
\end{scope}
\end{tikzpicture}
$$

### Inverse hyperbolic functions

Because $\sinh$ and $\tanh$ are one-to-one, and $\cosh$ becomes one-to-one when
restricted to $[0, \infty)$, they have inverses. These invert combinations of
exponentials, so their inverses are logarithms:
$$
\sinh^{-1} x = \ln\!\left(x + \sqrt{x^2 + 1}\right), \ x \in \mathbb{R},
\qquad
\cosh^{-1} x = \ln\!\left(x + \sqrt{x^2 - 1}\right), \ x \ge 1,
$$
$$
\tanh^{-1} x = \tfrac{1}{2}\ln\!\left(\frac{1 + x}{1 - x}\right), \ -1 < x < 1.
$$
The first is proved by solving $x = \sinh y = (e^y - e^{-y})/2$ as a quadratic
in $e^y$: multiplying by $e^y$ gives $(e^y)^2 - 2x\,e^y - 1 = 0$, so
$e^y = x + \sqrt{x^2 + 1}$ (the other root is negative and rejected), and taking
$\ln$ finishes it. Differentiating these logarithmic forms, or applying the
inverse-function rule, gives derivatives that are again purely algebraic:
$$
\frac{\d}{\d x}\sinh^{-1} x = \frac{1}{\sqrt{1 + x^2}},
\qquad
\frac{\d}{\d x}\cosh^{-1} x = \frac{1}{\sqrt{x^2 - 1}},
\qquad
\frac{\d}{\d x}\tanh^{-1} x = \frac{1}{1 - x^2}.
$$

Read backward, these give antiderivatives of radical and rational integrands,
the same forms that arise in
[trigonometric substitution](/calculus/techniques-of-integration/trigonometric-integrals-and-substitution).