Growth, Decay, Inverse Trigonometric, and Hyperbolic Functions

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 $d y / d t = k y$ , and the only functions solving that equation are exponentials. Solving it accounts for radioactive dating, cooling, and compound interest; the same inverse-function calculus then produces the inverse trigonometric and hyperbolic functions and their derivatives.

The law of natural growth

A solution is a function whose derivative is a constant multiple of itself — a description the natural exponential fits. Any $y (t) = C e^{k t}$ satisfies

y^{'} (t) = C k e^{k t} = k (C e^{k t}) = k y (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.

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 $d P / d t = 0.02 P$ with $t$ in years, the population grows at a relative rate of $2%$ per year and $P (t) = P_{0} e^{0.02 t}$ .

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).

Radioactive decay and half-life

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

Radiocarbon dating runs the same computation in reverse: measure the fraction of $^{14} C$ remaining and invert $m (t) / m_{0} = e^{k t}$ to recover the age.

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 (T - T_{s}),

where $T_{s}$ is the ambient temperature. The gap $y = T - T_{s}$ satisfies $y^{'} = k y$ because $T_{s}$ is constant, reducing the problem to natural decay, and $T (t) = T_{s} + (T (0) - T_{s}) e^{k t}$ .

Newton cooling. The temperature decays exponentially toward the ambient level Ts, the horizontal asymptote, from the initial gap T(0) - Ts.

Continuously compounded interest

An amount $A_{0}$ at annual rate $r$ compounded $n$ times a year grows to $A_{0} (1 + r / n)^{n t}$ after $t$ years. Letting $n \to \infty$ compounds continuously, and the limit definition of $e$ collapses the expression:

A (t) = n \to \infty lim A_{0} (1 + \frac{r}{n})^{n t} = A_{0} [m \to \infty lim (1 + \frac{1}{m})^{m}]^{r t} = A_{0} e^{r t},

with $m = n / r$ . Differentiating recovers $d A / d t = r A$ : continuous compounding is exactly the statement that the balance grows at a rate proportional to itself.

Process	Equation	Sign of $k$	Reported as
Population	$d P / d t = k P$	$k > 0$	relative growth rate
Radioactive decay	$d m / d t = k m$	$k < 0$	half-life
Newton cooling	$d T / d t = k (T - T_{s})$	$k < 0$	via gap $T - T_{s}$
Continuous interest	$d A / d t = r A$	$k = r > 0$	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 $[- π /2, π /2]$ increases from $- 1$ to $1$ , passes the Horizontal Line Test, and so has an inverse.

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

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.

Derivatives by implicit differentiation

Each derivative follows from the inverse-function method. Let $y = sin^{- 1} x$ , so $sin y = x$ with $y \in [- π /2, π /2]$ . Differentiating implicitly,

cos y \frac{d y}{d x} = 1 ⟹ \frac{d y}{d x} = \frac{1}{cos y} .

Since $y$ lies in $[- π /2, π /2]$ , $cos y \geq 0$ , so $cos y = 1 - sin^{2} y = 1 - x^{2}$ , giving an algebraic derivative with no trigonometry left in it:

\frac{d}{d x} sin^{- 1} x = \frac{1}{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 $b = 1 - x^{2}$ and $cos y = b = 1 - x^{2}$ . For arctangent, the legs are $1$ and $x$ , so the hypotenuse is $c = 1 + x^{2}$ and $sec y = c = 1 + x^{2}$ .

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).

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

The arctangent flattens toward horizontal asymptotes. Reflecting the vertical asymptotes of $tan x$ at $x = \pm π /2$ into the diagonal turns them into the horizontal lines $y = \pm π /2$ :

x \to \infty lim tan^{- 1} x = \frac{π}{2}, x \to - \infty lim tan^{- 1} x = - \frac{π}{2} .

The arctangent is defined for all x and levels off toward the two horizontal asymptotes, the mirror images of the tangent's vertical asymptotes.

Integrals from the inverse trig derivatives

Read backward, two of these formulas fill gaps in the integral table:

\int \frac{1}{1 - x ^{2}} d x = sin^{- 1} x + C, \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} (\frac{x}{a}) + C .

These are the antiderivatives that arise when integrating rational functions with irreducible quadratic denominators, a technique developed in partial fractions.

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.

Hyperbolic sine is odd with range $R$ ; 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.

The first identity is a direct computation and explains the name. Squaring the definitions,

cosh^{2} x - sinh^{2} x = \frac{e ^{2 x} + 2 + e ^{- 2 x}}{4} - \frac{e ^{2 x} - 2 + e ^{- 2 x}}{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 .

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 (\frac{x}{a}),

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.

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.

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 (x + x^{2} + 1), x \in R, cosh^{- 1} x = ln (x + x^{2} - 1), x \geq 1,

tanh^{- 1} x = \frac{1}{2} ln (\frac{1 + x}{1 - x}), - 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 $(e^{y})^{2} - 2 x e^{y} - 1 = 0$ , so $e^{y} = x + x^{2} + 1$ (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}{1 + x ^{2}}, \frac{d}{d x} cosh^{- 1} x = \frac{1}{x ^{2} - 1}, \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.