Inverse hyperbolic identities

Ask the tutor about Inverse hyperbolic identities Open the graphing workbench

Inverse hyperbolic functions already name the four principal branches and their $ln$ formulas. On those branches, how steep is the graph? And can one branch rewrite into another?

One step on arsinh

Drag the point on $y = arsinh x$ . The horizontal step has length $1$ . The vertical step equals $1/ 1 + x^{2}$ there, and the diagonal is the tangent. That is the same picture as one step on arcsin, except the radicand is $1 + x^{2}$ instead of $1 - x^{2}$ .

Drag the point on y = arsinh x. The horizontal step has length 1. The vertical step equals 1/√(1+x²) there, and the diagonal is the tangent.

For every real $x$ ,

$\frac{d}{d x} arsinh x = \frac{1}{1 + x ^{2}} .$

The other three slopes, each on its own domain

The same one-step reading on the remaining branches gives

$\frac{d}{d x} arcosh x = \frac{1}{x ^{2} - 1}, x > 1,$

$\frac{d}{d x} artanh x = \frac{1}{1 - x ^{2}}, ∣ x ∣ < 1,$

$\frac{d}{d x} arcoth x = \frac{1}{1 - x ^{2}}, ∣ x ∣ > 1.$

Artanh and arcoth share the same algebraic slope; the domains are complementary, matching the open intervals where each principal branch lives. Drag inside $(- 1, 1)$ . The horizontal step has length $1$ . The vertical step equals $1/ (1 - x^{2})$ . The dashed lines mark $x = \pm 1$ .

Drag the point on y = artanh x inside (−1, 1). The horizontal step has length 1. The vertical step equals 1/(1−x²) there. The dashed lines mark x = ±1.

Rewrite one branch into another

The four names are not four unrelated objects. For every real $x$ ,

$arsinh x = (sgn x) arcosh 1 + x^{2} = artanh \frac{x}{1 + x ^{2}} .$

(When $x = 0$ , the signed arcosh term is read as $0$ .) Drag a real $x$ . The green height is $asinh (x)$ . The blue height is $atanh (x / 1 + x^{2})$ . They stay equal.

Drag real x. The green height is asinh(x). The blue height is atanh(x / √(1+x²)). They stay equal.

For $x \geq 1$ ,

$arcosh x = arsinh x^{2} - 1 .$

For $∣ x ∣ > 1$ ,

$arcoth x = artanh \frac{1}{x} .$

Drag $x > 1$ . The green height is the closed form $\frac{1}{2} ln \frac{x + 1}{x - 1}$ of $arcoth x$ . The blue height is $atanh (1/ x)$ . They stay equal.

Drag x > 1. The green height is (1/2) ln((x+1)/(x−1)), the closed form of arcoth x. The blue height is atanh(1/x). They stay equal.

The slopes on the four principal branches, and the rewrites among them, are now in place. The next natural object in the handbook line is polynomials — a different family, not another inverse.