Part V · Complex Numbers, Trigonometry, and Advanced Functions

Chapter 16 — Rational, Inverse-Trigonometric, and Hyperbolic Functions

Why This Matters

Rational expressions appear in transfer-like functions and partial fractions. Inverse trigonometric functions recover angles from ratios. Hyperbolic functions arise naturally from exponentials and describe a geometry parallel to, but different from, circle trigonometry.

16.1 Function characteristics and the sigmoid

When a mathematical model is designed from requirements, begin with its desired characteristics: domain, range, symmetry, monotonicity, smoothness, limiting values, roots, and rate of change. The sigmoid

s(x)=\frac1{1+e^{-x}}

is a useful example. It is defined for every real $$x$$ , lies strictly between $$0$$ and $$1$$ , approaches $$0$$ as $x\to-\infty$ , approaches $$1$$ as $x\to\infty$ , and is strictly increasing because

s'(x)=\frac{e^{-x}}{(1+e^{-x})^2}>0.

It demonstrates a general design lesson: a function need not be chosen because it looks familiar. Its properties can be checked against the requested behaviour.

Invertibility requires a one-to-one mapping. A function $$f$$ has an inverse $f^{-1}$ on a domain if and only if it passes the Horizontal Line Test: each output corresponds to exactly one input.

Reflective symmetry across y = x

Starting point. Let $$(a, b)$$ lie on the graph of $$y = f(x)$$ , so $$b = f(a)$$ .

\[b = f(a) \iff a = f^{-1}(b)\]
By definition of the inverse function.reversible
\[(a, b) \in \text{graph}(f) \iff (b, a) \in \text{graph}(f^{-1})\]
Swapping inputs and outputs reflects every point across the diagonal line $$y = x$$ .reversible
\[(f^{-1})'(b) = \frac{1}{f'(a)} = \frac{1}{f'(f^{-1}(b))}\]
The slope of the reflected tangent line is the reciprocal of the original slope.reversible

Conclusion. $\text{Domain}(f^{-1}) = \text{Range}(f)$ and $\text{Range}(f^{-1}) = \text{Domain}(f)$ . The slopes are reciprocals: $(f^{-1})'(y) = \frac{1}{f'(x)}$ where $$y = f(x)$$ .

Check that $$f$$ is strictly monotonic (increasing or decreasing) on the specified interval.
Set $$y = f(x)$$ and solve algebraically for $$x$$ in terms of $$y$$ .
Swap variables: write $y = f^{-1}(x)$ .
State domain and range restrictions explicitly.

16.2 Rational functions and partial fractions

A rational function is $$N(x)/D(x)$$ where $$N$$ and $$D$$ are polynomials and $D(x)\ne0$ . Zeros come from roots of $$N$$ not cancelled by $$D$$ ; roots of $$D$$ give excluded inputs and may give poles. If the numerator degree is at least the denominator degree, perform polynomial division first.

For distinct linear factors,

\frac{P(x)}{(x-a)(x-b)}=\frac A{x-a}+\frac B{x-b}.

Multiplying through by the denominator and choosing convenient $$x$$ values finds the coefficients. Repeated factors and irreducible quadratics require expanded templates; do not use the distinct-linear template blindly.

Example — partial fractions and a local power-series view

Decompose

\frac1{x^2-1}=\frac A{x-1}+\frac B{x+1}.

Multiplication by $$(x-1)(x+1)$$ gives

\[1=A(x+1)+B(x-1).\]

Putting $$x=1$$ gives $$A=1/2$$ ; putting $$x=-1$$ gives $$B=-1/2$$ . Hence

\frac1{x^2-1}=\frac1{2(x-1)}-\frac1{2(x+1)}.

Partial fractions reveal both poles, $x=\pm1$ . They can also prepare a power series. For example, when $$|x|<1$$ ,

\frac1{1-x}=1+x+x^2+\cdots,

because it is the geometric series with ratio $$x$$ . The inequality $$|x|<1$$ is part of the statement; a rational expression exists at many values where this particular series does not converge.

Decomposition reverses recombination. A rational expression with factored denominator can be written as simpler fractions whose common-denominator sum reproduces the original numerator. Solving for the constants is coefficient matching: equality for every permitted input forces matching polynomial coefficients.

Factor the denominator and write the required fraction shape.
Clear denominators to obtain a polynomial identity.
Choose convenient inputs or match coefficients, then verify by recombining.

16.3 Inverse trigonometric functions

Trigonometric functions are periodic, so they must be restricted before they can have inverse functions. Standard principal branches are

\arcsin:[-1,1]\to[-\pi/2,\pi/2],\quad \arccos:[-1,1]\to[0,\pi],

\arctan:\mathbb R\to(-\pi/2,\pi/2).

The notation $\arcsin x$ avoids confusing an inverse with $1/\sin x$ .

Implicit differentiation of $\sin y=x$ gives

\cos y\frac{dy}{dx}=1.

On the arcsine range, $\cos y\ge0$ , and $\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2}$ . Thus

\frac d{dx}\arcsin x=\frac1{\sqrt{1-x^2}},\qquad -1<x<1.

Similarly,

\frac d{dx}\arccos x=-\frac1{\sqrt{1-x^2}},\qquad -1<x<1,

\frac d{dx}\arctan x=\frac1{1+x^2}\qquad(x\in\mathbb R).

For $$a>0$$ ,

\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\frac xa+C, \qquad \int\frac{dx}{a^2+x^2}=\frac1a\arctan\frac xa+C.

The reciprocal inverse functions require especially careful absolute values. On their standard principal branches,

\frac{d}{dx}\operatorname{arcsec}x= \frac{1}{|x|\sqrt{x^2-1}},\qquad \frac{d}{dx}\operatorname{arccsc}x= -\frac{1}{|x|\sqrt{x^2-1}},\qquad |x|>1.

The $$|x|$$ is not optional: $\sqrt{x^2}=|x|$ , not generally $$x$$ . Different texts make different endpoint choices for these inverse functions; the displayed derivatives are asserted only on intervals where a consistent branch has been chosen.

Why a branch is necessary. Sine repeats values, so it has no inverse on all real inputs. Restricting it to one interval makes each allowed output identify exactly one angle; the inverse notation records that chosen branch.

Restricted sine branch and its inverse reflection in y equals x.

16.4 Hyperbolic functions

Define

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

$\cosh x$ is even and $\sinh x$ is odd. Their identity is

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

so $(\cosh t,\sinh t)$ traces the hyperbola $$X^2-Y^2=1$$ , unlike the circle identity for sine and cosine. Their derivatives are

(\cosh x)'=\sinh x,\qquad(\sinh x)'=\cosh x.

The inverse functions have logarithmic forms on their principal real domains:

\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right),

\operatorname{arcosh}x=\ln\left(x+\sqrt{x^2-1}\right),\qquad x\ge1.

The quotient

\tanh x=\frac{\sinh x}{\cosh x} =\frac{e^x-e^{-x}}{e^x+e^{-x}}

is defined for every real $$x$$ , is odd, and satisfies $-1<\tanh x<1$ . The identity $\cosh^2x-\sinh^2x=1$ gives

(\tanh x)'=\frac{\cosh^2x-\sinh^2x}{\cosh^2x} =\operatorname{sech}^2x, \qquad \operatorname{sech}x=\frac1{\cosh x}.

Its inverse has the real domain $$-1<x<1$$ :

\operatorname{artanh}x=\frac12\ln\left(\frac{1+x}{1-x}\right), \qquad \frac{d}{dx}\operatorname{artanh}x=\frac1{1-x^2}.

The remaining useful inverse-hyperbolic derivatives are

\frac{d}{dx}\operatorname{arsinh}x=\frac1{\sqrt{1+x^2}} \quad(x\in\mathbb R), \qquad \frac{d}{dx}\operatorname{arcosh}x=\frac1{\sqrt{x^2-1}} \quad(x>1).

Consequently, for $$a>0$$ ,

\int\frac{dx}{\sqrt{a^2+x^2}} =\operatorname{arsinh}\frac xa+C, \qquad \int\frac{dx}{a^2-x^2} =\frac1a\operatorname{artanh}\frac xa+C

on intervals where the integrands are defined. The latter statement is local: its denominator excludes $x=\pm a$ .

Build the hyperbolic functions from exponentials. The symmetric combination keeps the even part of the exponential, while the antisymmetric combination keeps the odd part. Adding and subtracting their definitions reconstructs e^x and e^{-x}, which is the quickest route to many identities.

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

The even cosh curve and odd sinh curve built from exponential components.

16.5 Inverse functions: branch first, formula second

An inverse function must return one output for each permitted input. Because $\sin\theta=1/2$ at infinitely many angles, $\arcsin(1/2)$ cannot mean “all solutions.” It means the one angle in the chosen interval $[-\pi/2,\pi/2]$ , namely $\pi/6$ . Solving a trigonometric equation is a different task: after finding a reference angle, all periodic solutions must be listed if the problem requests them.

[Figure: Unit circle with the horizontal line

\(y=1/2\)

. Mark both angles whose sine is

\(1/2\)

, then highlight

\pi/6

as the value chosen by the arcsine principal branch.]

Example — inverse value versus all angles

The statement

\arccos(-1/2)=\frac{2\pi}{3}

uses the range $[0,\pi]$ . In contrast, the equation $\cos\theta=-1/2$ has

\theta=\frac{2\pi}{3}+2\pi k \quad\hbox{or}\quad \theta=\frac{4\pi}{3}+2\pi k,\qquad k\in\mathbb Z.

Mixing these two answers is a branch error, not an arithmetic error.

Example — derive the arctangent derivative

Let $y=\arctan x$ , so $\tan y=x$ with $-\pi/2<y<\pi/2$ . Differentiate:

\sec^2y\frac{dy}{dx}=1.

The identity $\sec^2y=1+\tan^2y=1+x^2$ gives

\frac{dy}{dx}=\frac1{1+x^2}.

Unlike arcsine, no square-root sign choice is needed because the tangent identity already produces a positive denominator. The formula is valid for all real $$x$$ .

Calculus of hyperbolic and inverse hyperbolic functions. Hyperbolic derivatives mirror trigonometric derivatives, but with key sign changes due to $\cosh^2 x - \sinh^2 x = 1$ .

Hyperbolic derivatives

Starting point. Differentiate the exponential definitions of $\sinh x$ and $\cosh x$ .

\[\frac{d}{dx}\sinh x = \frac{d}{dx}\left(\frac{e^x - e^{-x}}{2}\right) = \frac{e^x - (-e^{-x})}{2} = \frac{e^x + e^{-x}}{2} = \cosh x\]
The derivative of $\sinh x$ is $+\cosh x$ .reversible
\[\frac{d}{dx}\cosh x = \frac{d}{dx}\left(\frac{e^x + e^{-x}}{2}\right) = \frac{e^x - e^{-x}}{2} = \sinh x\]
The derivative of $\cosh x$ is $+\sinh x$ (positive sign, unlike $-\sin x$ ).reversible
\[\frac{d}{dx}\tanh x = \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} = \frac{1}{\cosh^2 x} = \operatorname{sech}^2 x\]
Apply the quotient rule and use $\cosh^2 x - \sinh^2 x = 1$ .reversible

Conclusion. $(\sinh x)' = \cosh x$ , $(\cosh x)' = \sinh x$ , $(\tanh x)' = \operatorname{sech}^2 x$ . Inverse hyperbolic formulas: $\operatorname{arsinh} x = \ln(x + \sqrt{x^2 + 1})$ , $\operatorname{arcosh} x = \ln(x + \sqrt{x^2 - 1})$ ( $x \ge 1$ ).

$\int \sinh x\,dx = \cosh x + C$ , $\int \cosh x\,dx = \sinh x + C$ .
$\int \frac{dx}{\sqrt{x^2 + a^2}} = \operatorname{arsinh}\left(\frac{x}{a}\right) + C = \ln\left(x + \sqrt{x^2 + a^2}\right) + C$ .
$\int \frac{dx}{\sqrt{x^2 - a^2}} = \operatorname{arcosh}\left(\frac{x}{a}\right) + C = \ln\left(x + \sqrt{x^2 - a^2}\right) + C$ ( $$x > a > 0$$ ).

16.6 Inverse-trigonometric integrals by pattern recognition

The expressions $$a^2-x^2$$ and $$a^2+x^2$$ signal different geometries. Before using a formula, factor constants until the denominator exactly matches one of the standard patterns.

Example — an arcsine pattern

Evaluate

\int\frac{dx}{\sqrt{25-4x^2}}.

Factor $$25$$ inside the square root:

\sqrt{25-4x^2}=5\sqrt{1-\left(\frac{2x}{5}\right)^2}.

Let $$u=2x/5$$ , so $dx=(5/2)\,du$ . Then

\int\frac{dx}{\sqrt{25-4x^2}} =\frac12\int\frac{du}{\sqrt{1-u^2}} =\frac12\arcsin u+C =\frac12\arcsin\frac{2x}{5}+C.

For real values, the original integrand requires $$|x|<5/2$$ on an interval; at the endpoints the derivative form is unbounded.

Example — an arctangent pattern

Evaluate

\int\frac{3\,dx}{x^2+9}.

Since $$x^2+9=x^2+3^2$$ ,

\int\frac{3\,dx}{x^2+9} =3\left(\frac13\arctan\frac{x}{3}\right)+C =\arctan\frac{x}{3}+C.

Differentiating $\arctan(x/3)$ gives $$(1/3)/(1+x^2/9)=3/(x^2+9)$$ , which checks both the scale and the result.

Inverse trigonometric functions require restricted principal branches. Because $\sin$ , $\cos$ , and $\tan$ are periodic, their domains must be restricted to establish one-to-one branches.

Derivatives of inverse trigonometric functions

Starting point. Let $y = \arcsin x$ for $x \in [-1, 1]$ and $y \in [-\pi/2, \pi/2]$ .

\[\sin y = x \implies \frac{d}{dx}(\sin y) = \frac{d}{dx}(x) \implies \cos y \frac{dy}{dx} = 1\]
Differentiate implicitly with respect to $$x$$ .reversible
\[\frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - \sin^2 y}} = \frac{1}{\sqrt{1 - x^2}}\]
Since $y \in [-\pi/2, \pi/2]$ , $\cos y \ge 0$ , so $\cos y = +\sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$ .reversible

Conclusion. $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}$ , $\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^2}}$ , $\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}$ .

$\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\left(\frac{x}{a}\right) + C$ ( $$a > 0$$ ).
$\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$ ( $$a > 0$$ ).
Look for quadratic denominators: complete the square to fit these standard inverse-trig forms.

16.7 Hyperbolic functions are exponential combinations

Circular functions use a plus sign in $\sin^2x+\cos^2x=1$ and are periodic. Hyperbolic functions use a minus sign in $\cosh^2x-\sinh^2x=1$ and are not periodic. From their exponential definitions,

\cosh x+\sinh x=e^x,\qquad \cosh x-\sinh x=e^{-x}.

Adding and subtracting those equations recovers the definitions. This makes hyperbolic functions natural whenever a calculation already contains both growing and decaying exponentials.

Example — a hyperbolic simplification

Show that $\tanh x$ approaches $$1$$ as $x\to\infty$ . Divide numerator and denominator by $$e^x$$ :

\tanh x=\frac{1-e^{-2x}}{1+e^{-2x}}\longrightarrow\frac{1-0}{1+0}=1.

Similarly it approaches $$-1$$ as $x\to-\infty$ . The bounded S-shape contrasts with $\tan x$ , which has vertical asymptotes.

Example — inverse-hyperbolic integral

Evaluate

\int\frac{dx}{\sqrt{x^2+16}}.

The form is $$a^2+x^2$$ , but it is inside a square root rather than an ordinary quadratic denominator. With $$a=4$$ ,

\int\frac{dx}{\sqrt{x^2+16}} =\operatorname{arsinh}\frac{x}{4}+C =\ln\left(\frac{x+\sqrt{x^2+16}}{4}\right)+C.

The two antiderivatives differ from $\ln(x+\sqrt{x^2+16})$ only by the constant $-\ln4$ , so all are equivalent indefinite answers.

Hyperbolic functions parametrize hyperbolas. Just as $(\cos t, \sin t)$ parametrizes the circle $$x^2 + y^2 = 1$$ , the hyperbolic pair $(\cosh t, \sinh t)$ parametrizes the hyperbola $$x^2 - y^2 = 1$$ .

Definitions and fundamental hyperbolic identity

Starting point. Define $\cosh x = \frac{e^x + e^{-x}}{2}$ (even part of $$e^x$$ ) and $\sinh x = \frac{e^x - e^{-x}}{2}$ (odd part of $$e^x$$ ).

\[\cosh^2 x = \frac{(e^x + e^{-x})^2}{4} = \frac{e^{2x} + 2 + e^{-2x}}{4}\]
Square $\cosh x$ algebraically.reversible
\[\sinh^2 x = \frac{(e^x - e^{-x})^2}{4} = \frac{e^{2x} - 2 + e^{-2x}}{4}\]
Square $\sinh x$ algebraically.reversible
\[\cosh^2 x - \sinh^2 x = \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} = \frac{4}{4} = 1\]
Subtract: the exponential terms cancel, leaving identically $$1$$ .reversible

Conclusion. $\cosh^2 x - \sinh^2 x = 1$ for all real $$x$$ . Also: $\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$ .

Common Mistakes

Cancelling a rational factor and forgetting its excluded input.
Using inverse trig notation without a specified principal range.
Losing the positive square-root sign in the arcsine derivative.
Replacing the hyperbolic identity with the circular identity; its sign differs.

Exercises

Decompose $$1/(x^2-1)$$ into partial fractions.
Find $\arcsin(\sqrt3/2)$ and explain the principal-value choice.
Differentiate $\arctan(2x)$ .
Verify $\cosh^2x-\sinh^2x=1$ from exponential definitions.
Sketch the qualitative difference between $\tanh x$ and $\tan x$ .
Differentiate $\operatorname{arccos}(x)$ and explain why its sign differs from that of $\operatorname{arcsin}(x)$ .
Evaluate $\int dx/\sqrt{9+x^2}$ and state why an inverse hyperbolic function is appropriate.
Distinguish $\arcsin(1/2)$ from all solutions of $\sin\theta=1/2$ .
Evaluate $\int dx/\sqrt{16-9x^2}$ and state the real domain of the integrand.
Evaluate $\int 5\,dx/(x^2+25)$ .
Prove from definitions that $\cosh x+\sinh x=e^x$ .
Sketch the key qualitative differences between $\tan x$ and $\tanh x$ .