Inverse hyperbolic rule · Derivative Calculator

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]

From $\sinh y = x$.
$\cosh y\cdot y' = 1$
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]

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]

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]

Chain rule with $u = 2x$.
$= \frac{1}{1 - \left(2x\right)^{2}}\cdot 2$
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]

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 rule Inverse trigonometric rule Chain rule

What it says

When it applies

Five worked examples

Watch out for

Related rules