Inverse Hyperbolic Functions

Inverse hyperbolic functions are the inverse functions of the hyperbolic sine, cosine, tangent, and other hyperbolic functions. They are used to solve equations involving hyperbolic functions and are expressed in terms of logarithmic formulas. 

The three basic inverse hyperbolic functions are:

• Inverse hyperbolic sine (sinh^-1 or arcsinh)
• Inverse hyperbolic cosine (cosh^-1 or arccosh)
• Inverse hyperbolic tangent (tanh^-1 or arctanh)

These functions, similar to the trigonometric (or circular) functions, are widely applied in mathematics for solving integrals and differential equations.

Inverse Hyperbolic Sine (sinh^-1)

The formula of the inverse hyperbolic sine is given by:

sinh^-1 x = ${\ln \left( x+\sqrt{1+x^{2}}\right)}$

Proof

Let sinh^-1 x = y, where y ∈ ℝ

⇒ x = sinh y

Using the exponential combination of the hyperbolic sine function, we get

⇒ sinh y = ${\dfrac{e^{y}-e^{-y}}{2}}$

⇒ x = ${\dfrac{e^{y}-e^{-y}}{2}}$

⇒ 2x = e^y – e^-y 

⇒ e^2y – 2xe^y – 1 = 0 …..(i)

Here, equation (i) is of the form ax² + bx + c = 0 …..(ii)

Since the roots of equation (ii) are 

x = ${\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

Now, from the equation (i), we get 

e^y = ${\dfrac{-\left( -2x\right) \pm \sqrt{\left( -2x\right) ^{2}-4\left( 1\right) \left( -1\right) }}{2\left( 1\right) }}$

= ${\dfrac{2x\pm \sqrt{4\left( x^{2}+1\right) }}{2}}$

= ${\dfrac{2x\pm 2\sqrt{\left( x^{2}+1\right) }}{2}}$

= ${x\pm \sqrt{\left( x^{2}+1\right) }}$

Since y ∈ ℝ, e must be a positive number (i.e., e > 0).

e^y = ${x+\sqrt{\left( x^{2}+1\right) }}$

⇒ y = ${ln\left( x+\sqrt{\left( x^{2}+1\right) }\right)}$

⇒ sinh^-1 x = ${ln\left( x+\sqrt{\left( x^{2}+1\right) }\right)}$

Thus, sinh^-1 x = ${\ln \left( x+\sqrt{1+x^{2}}\right)}$ 

On graphing the inverse hyperbolic sine function y = sinh^-1 x, we get:

Inverse Hyperbolic Cosine (cosh^-1)

The formula of the inverse hyperbolic cosine is given by:

cosh^-1 x = ${\ln \left( x+\sqrt{x^{2}-1}\right)}$

Proof

Let cosh^-1 x = y, where y ∈ ℝ

⇒ x = cosh y

Using the exponential combination of the hyperbolic cosine function, we get

⇒ cosh y = ${\dfrac{e^{y}+e^{-y}}{2}}$

⇒ x = ${\dfrac{e^{y}+e^{-y}}{2}}$

⇒ 2x = e^y + e^-y 

⇒ e^2y – 2xe^y + 1 = 0 …..(i)

Here, the equation (i) is of the form ax² + bx + c = 0 …..(ii)

Since the roots of equation (ii) are 

x = ${\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

Now, from the equation (i), we get 

e^y = ${\dfrac{-\left( -2x\right) \pm \sqrt{\left( -2x\right) ^{2}-4\left( 1\right) \left( 1\right) }}{2\left( 1\right) }}$

= ${\dfrac{2x\pm \sqrt{4\left( x^{2}-1\right) }}{2}}$

= ${\dfrac{2x\pm 2\sqrt{\left( x^{2}-1\right) }}{2}}$

= ${x\pm \sqrt{\left( x^{2}-1\right) }}$

Since y ∈ ℝ, e must be a positive number (i.e., e > 0).

e^y = ${x+\sqrt{\left( x^{2}-1\right) }}$

⇒ y = ${ln\left( x+\sqrt{\left( x^{2}-1\right) }\right)}$

⇒ cosh^-1 x = ${ln\left( x+\sqrt{\left( x^{2}-1\right) }\right)}$

Thus, cosh^-1 x = ${\ln \left( x+\sqrt{x^{2}-1}\right)}$, here x ≥ 1

On graphing the inverse hyperbolic cosine function y = cosh^-1 x, we get:

Inverse Hyperbolic Tangent (tanh^-1)

The formula of the inverse hyperbolic tangent is given by:

tanh^-1 x = ${\dfrac{1}{2}\ln \left( \dfrac{1+x}{1-x}\right)}$

Proof

Let tanh^-1 x = y, where y ∈ ℝ

⇒ x = tanh y

Using the exponential combination of the hyperbolic tangent function, we get

tanh y = ${\dfrac{e^{y}-e^{-y}}{e^{y}+e^{-y}}}$

⇒ x = ${\left( \dfrac{e^{y}-e^{-y}}{e^{y}+e^{-y}}\right) \times \left( \dfrac{e^{y}}{e^{y}}\right)}$

⇒ x = ${\dfrac{e^{2y}-1}{e^{2y}+1}}$

⇒ x(e^2y + 1) = (e^2y – 1)

⇒ (x – 1)e^2y + (x + 1) = 0

⇒ e^2y = ${-\dfrac{x+1}{x-1}}$

⇒ e^2y = ${\dfrac{1+x}{1-x}}$

⇒ 2y = ${\ln \left( \dfrac{1+x}{1-x}\right)}$

⇒ y = ${\dfrac{1}{2}\ln \left( \dfrac{1+x}{1-x}\right)}$

⇒ tanh^-1 x = ${\dfrac{1}{2}\ln \left( \dfrac{1+x}{1-x}\right)}$

Thus, tanh^-1 x = ${\dfrac{1}{2}\ln \left( \dfrac{1+x}{1-x}\right)}$, here, |x| < 1

On graphing the inverse hyperbolic tangent function y = tanh^-1 x, we get:

Other Inverse Hyperbolic Functions

In addition to the three basic inverse hyperbolic functions, there are three other inverse hyperbolic functions corresponding to sine, cosine, and tangent. 

Inverse Hyperbolic Cosecant (cosech^-1 or csch^-1)

The formula of the inverse hyperbolic cosecant is expressed as:

cosech^-1 x = ${\ln \left( \dfrac{1+\sqrt{1+x^{2}}}{x}\right)}$

Proof

Let csch^-1 x = y, where y ∈ ℝ

⇒ x = csch y

Using the exponential combination of the hyperbolic cosecant function, we get

csch y or cosech y = ${\dfrac{1}{\sinh y}}$ = ${\dfrac{2}{e^{y}-e^{-y}}}$

⇒ x = ${\dfrac{2}{e^{y}-e^{-y}}}$

⇒ x = ${\left( \dfrac{2}{e^{y}-e^{-y}}\right) \times \left( \dfrac{e^{y}}{e^{y}}\right)}$

⇒ x = ${\dfrac{2e^{y}}{e^{2y}-1}}$

⇒ x(e^2y – 1) = 2e^y

⇒  xe^2y – 2e^y – x = 0 …..(i)

Here, the equation (i) is of the form ax² + bx + c = 0 …..(ii)

Since the roots of equation (ii) are 

x = ${\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

Now, from the equation (i), we get 

⇒ e^y = ${\dfrac{-\left( -2\right) \pm \sqrt{\left( -2\right) ^{2}-4\left( x\right) \left( -x\right) }}{2\left( x\right) }}$

⇒ e^y = ${\dfrac{2\pm \sqrt{4+4x^{2}}}{2x}}$

⇒ e^y = ${\dfrac{2\pm \sqrt{4\left( 1+x^{2}\right) }}{2x}}$

⇒ e^y = ${\dfrac{2\pm 2\sqrt{1+x^{2}}}{2x}}$

⇒ e^y = ${\dfrac{1\pm \sqrt{1+x^{2}}}{x}}$

Since y ∈ ℝ, e must be a positive number (i.e., e > 0).

e^y = ${\dfrac{1+\sqrt{1+x^{2}}}{x}}$

⇒ y = ${\ln \dfrac{1+\sqrt{1+x^{2}}}{x}}$

⇒ csch^-1 x = ${\ln \dfrac{1+\sqrt{1+x^{2}}}{x}}$

⇒ csch^-1 x = ${\ln \dfrac{1+\sqrt{1+x^{2}}}{x}}$

Thus, csch^-1 x = ${\ln \left( \dfrac{1+\sqrt{1+x^{2}}}{x}\right)}$, here, x ≠ 0

On graphing the inverse hyperbolic cosecant function y = csch^-1 x, we get:

Inverse Hyperbolic Secant (sech^-1)

The formula of the inverse hyperbolic secant is written as:

sech^-1 x = ${\ln \left( \dfrac{1+\sqrt{1-x^{2}}}{x}\right)}$

Proof

Let sech^-1 x = y, where y ∈ ℝ

⇒ x = sech y

Using the exponential combination of the hyperbolic secant function, we get

sech y = ${\dfrac{1}{\cosh y}}$ = ${\dfrac{2}{e^{y}+e^{-y}}}$ 

⇒ x = ${\dfrac{2}{e^{y}+e^{-y}}}$

⇒ x = ${\left( \dfrac{2}{e^{y}+e^{-y}}\right) \times \left( \dfrac{e^{y}}{e^{y}}\right)}$

⇒ x = ${\dfrac{2e^{y}}{e^{2y}+1}}$

⇒ x(e^2y + 1) = 2e^y

⇒  xe^2y – 2e^y + x = 0 …..(i)

Here, the equation (i) is of the form ax² + bx + c = 0 …..(ii)

Since the roots of equation (ii) are 

x = ${\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

Now, from the equation (i), we get 

⇒ e^y = ${\dfrac{-\left( -2\right) \pm \sqrt{\left( -2\right) ^{2}-4\left( x\right) \left( x\right) }}{2\left( x\right) }}$

⇒ e^y = ${\dfrac{2\pm \sqrt{4-4x^{2}}}{2x}}$

⇒ e^y = ${\dfrac{2\pm \sqrt{4\left( 1-x^{2}\right) }}{2x}}$

⇒ e^y = ${\dfrac{2\pm 2\sqrt{1-x^{2}}}{2x}}$

⇒ e^y = ${\dfrac{1\pm \sqrt{1-x^{2}}}{x}}$

Since y ∈ ℝ, e must be a positive number (i.e., e > 0).

e^y = ${\dfrac{1+\sqrt{1-x^{2}}}{x}}$

⇒ y = ${\ln \dfrac{1+\sqrt{1-x^{2}}}{x}}$

⇒ sech^-1 x = ${\ln \dfrac{1+\sqrt{1-x^{2}}}{x}}$

Thus, sech^-1 x = ${\ln \left( \dfrac{1+\sqrt{1-x^{2}}}{x}\right)}$, here, 0 < x ≤ 1

On graphing the inverse hyperbolic secant function y = sech^-1 x, we get:

Inverse Hyperbolic Cotangent (coth^-1)

The formula of the inverse hyperbolic cotangent is expressed as:

coth^-1 x = ${\dfrac{1}{2}\ln \left( \dfrac{x+1}{x-1}\right)}$ 

Proof

Let coth¹ x = y, where y ∈ ℝ

⇒ x = coth y

Using the exponential combination of the hyperbolic cotangent function, we get

coth y = ${\dfrac{1}{\tanh y}}$ = ${\dfrac{e^{y}+e^{-y}}{e^{y}-e^{-y}}}$

⇒ x = ${\dfrac{e^{y}+e^{-y}}{e^{y}-e^{-y}}}$

⇒ x = ${\left( \dfrac{e^{y}+e^{-y}}{e^{y}-e^{-y}}\right) \times \left( \dfrac{e^{y}}{e^{y}}\right)}$

⇒ x = ${\dfrac{e^{2y}+1}{e^{2y}-1}}$

⇒ x(e^2y – 1) = (e^2y + 1)

⇒ (x – 1)e^2y – (x + 1) = 0

⇒ e^2y = ${\dfrac{x+1}{x-1}}$

⇒ 2y = ${\ln \left( \dfrac{x+1}{x-1}\right)}$

⇒ y = ${\dfrac{1}{2}\ln \left( \dfrac{x+1}{x-1}\right)}$

⇒ coth¹ x = ${\dfrac{1}{2}\ln \left( \dfrac{x+1}{x-1}\right)}$

Thus, coth¹ x = ${\dfrac{1}{2}\ln \left( \dfrac{x+1}{x-1}\right)}$, here, |x| > 1

On graphing the inverse hyperbolic cotangent function y = coth^-1 x, we get:

Here is the summary of all graphs of the inverse hyperbolic functions:

Each inverse hyperbolic function has a different domain and a range that should be considered. The domains and ranges of these functions are shown in the following table:

Inverse Hyperbolic Function	Domain	Range
sinh-1 x	(-∞, ∞)	(-∞, ∞)
cosh-1 x	[1, ∞)	[0, ∞)
tanh-1 x	(-1, 1)	(-∞, ∞)
cosech-1 x	(-∞, ∞)	(-∞, ∞)
sech-1 x	(0, 1]	[0, ∞)
coth-1 x	(-∞, -1) ∪ (1, ∞)	(-∞, ∞)

Derivatives

Here are the derivatives of the inverse hyperbolic functions:

Inverse Hyperbolic Function	Formula With Restriction
${\dfrac{d}{dx}\left( arcsinh \  x\right)}$	${\dfrac{1}{\sqrt{1+x^{2}}}}$
${\dfrac{d}{dx}\left( arccosh \  x\right)}$	${\dfrac{1}{\sqrt{x^{2}-1}}}$, here x > 1
${\dfrac{d}{dx}\left( arctanh \  x\right)}$	${\dfrac{1}{1-x^{2}}}$, here |x| < 1
${\dfrac{d}{dx}\left( arccosech \  x\right)}$	${-\dfrac{1}{\left| x\right| \sqrt{1-x^{2} }}}$, here x ≠ 0
${\dfrac{d}{dx}\left( arcsech \  x\right)}$	${-\dfrac{1}{x\sqrt{1-x^{2} }}}$, here 0 < x < 1
${\dfrac{d}{dx}\left( arccoth \  x\right)}$	${\dfrac{1}{1-x^{2}}}$, here |x| > 1

Solved Examples