Derivative Calculator

Inverse hyperbolic rule

Like the inverse trigonometric derivatives, but with the signs inside the roots flipped.

\[\frac{d}{dx}\operatorname{artanh} x = \frac{1}{1 - x^{2}}\]

What it says

The derivation is the same inverse-function argument. For \(y = \operatorname{arsinh} x\), \(\sinh y = x\) gives \(\cosh y\cdot y' = 1\), and since \(\cosh y = \sqrt{1 + \sinh^{2}y} = \sqrt{1 + x^{2}}\), the derivative is \(\frac{1}{\sqrt{x^{2} + 1}}\).

The table: \(\left(\operatorname{arsinh} x\right)' = \frac{1}{\sqrt{x^{2} + 1}}\), \(\left(\operatorname{arcosh} x\right)' = \frac{1}{\sqrt{x^{2} - 1}}\) for \(x > 1\), \(\left(\operatorname{artanh} x\right)' = \frac{1}{1 - x^{2}}\) for \(\left|x\right| < 1\).

Compare with the inverse trigonometric versions: \(1 - x^{2}\) becomes \(x^{2} \pm 1\), and \(1 + x^{2}\) becomes \(1 - x^{2}\). Every difference is a sign.

When it applies

  • \(\operatorname{arsinh}\), \(\operatorname{arcosh}\), \(\operatorname{artanh}\), written here as \(\texttt{asinh}\), \(\texttt{acosh}\), \(\texttt{atanh}\).
  • Integrals of \(\frac{1}{\sqrt{x^{2} \pm 1}}\), read backwards.
  • With the chain rule for compound arguments.

Five worked examples

Every line is the step the calculator would show, in the order it applies them. Each graph is live: hover it to read both curves and see the tangent whose slope is the derivative, drag to pan, scroll to zoom.

Example 1
\[\frac{d}{dx}\left[\operatorname{arsinh}\left(x\right)\right]\]

  1. From \(\sinh y = x\).

    \[\cosh y\cdot y' = 1\]

  2. Substitute \(\cosh y = \sqrt{1 + x^{2}}\).

    \[y' = \frac{1}{\sqrt{x^{2} + 1}}\]
Answer
\[\frac{1}{\sqrt{x^{2} + 1}}\]

f is a gentle S through the origin, growing like a logarithm rather than levelling off. f′ is a bump of height 1 that decays in both directions but never quite reaches zero.

Defined for every real \(x\), since \(x^{2} + 1\) is never zero.

Open asinh(x) in the calculator
Example 2
\[\frac{d}{dx}\left[\operatorname{arcosh}\left(x\right)\right]\]

  1. From \(\cosh y = x\), using \(\sinh y = \sqrt{x^{2} - 1}\).

    \[y' = \frac{1}{\sqrt{x^{2} - 1}}\]
Answer
\[\frac{1}{\sqrt{x^{2} - 1}}\]

The curve begins at \(x = 1\) with a vertical tangent, which is why f′ blows up there before decaying.

Only for \(x > 1\) — below that the inverse does not exist.

Open acosh(x) in the calculator
Example 3
\[\frac{d}{dx}\left[\operatorname{artanh}\left(x\right)\right]\]

  1. From \(\tanh y = x\), with \(\operatorname{sech}^{2}y = 1 - x^{2}\).

    \[y' = \frac{1}{1 - x^{2}}\]
Answer
\[\frac{1}{1 - x^{2}}\]

Confined to \((-1, 1)\). f runs off to \(\pm\infty\) at the two ends and f′ blows up with it, bottoming out at 1.

Compare \(\left(\arctan x\right)' = \frac{1}{1 + x^{2}}\): the same expression with the sign reversed.

Open atanh(x) in the calculator
Example 4
\[\frac{d}{dx}\left[\operatorname{artanh}\left(2 x\right)\right]\]

  1. Chain rule with \(u = 2x\).

    \[= \frac{1}{1 - \left(2x\right)^{2}}\cdot 2\]

  2. Square the inside.

    \[= \frac{2}{1 - 4x^{2}}\]
Answer
\[\frac{2}{1 - 4 x^{2}}\]

The window is barely half a unit wide, and f′ is a steep-walled well with its floor at 2 above the origin.

Valid on \(\left|x\right| < \tfrac{1}{2}\), where the argument stays inside \((-1, 1)\).

Open atanh(2x) in the calculator
Example 5
\[\frac{d}{dx}\left[x \operatorname{arsinh}\left(x\right)\right]\]

  1. Product rule.

    \[= 1\cdot\operatorname{arsinh} x + \frac{x}{\sqrt{x^{2} + 1}}\]
Answer
\[\operatorname{arsinh}\left(x\right) + \frac{x}{\sqrt{x^{2} + 1}}\]

A shallow valley whose floor is the origin, where f′ changes sign.

The pairing behind \(\int \operatorname{arsinh} x\,dx = x\operatorname{arsinh} x - \sqrt{x^{2} + 1}\).

Open x asinh(x) in the calculator

Watch out for

  • The signs under the roots differ between \(\operatorname{arsinh}\) \(\left(x^{2} + 1\right)\) and \(\operatorname{arcosh}\) \(\left(x^{2} - 1\right)\).
  • \(\operatorname{arcosh}\) needs \(x > 1\) and \(\operatorname{artanh}\) needs \(\left|x\right| < 1\); outside those ranges the derivative is meaningless.
  • Unlike \(\arccos\), \(\operatorname{arcosh}\) has a positive derivative.
Hyperbolic ruleInverse trigonometric ruleChain rule