# Explain Inverse Hyperbolic Functions Formula

In mathematics, the inverse functions of hyperbolic functions are referred to as inverse hyperbolic functions or area hyperbolic functions. There are six inverse hyperbolic functions, namely, inverse hyperbolic sine, inverse hyperbolic cosine, inverse hyperbolic tangent, inverse hyperbolic cosecant, inverse hyperbolic secant, and inverse hyperbolic cotangent functions. These functions are depicted as sinh^{-1} x, cosh^{-1 }x, tanh^{-1} x, csch^{-1} x, sech^{-1} x, and coth^{-1} x. With the help of an inverse hyperbolic function, we can find the hyperbolic angle of the corresponding hyperbolic function.

Function name |
Function |
Formula |
Domain |
Range |
---|---|---|---|---|

Inverse hyperbolic sine |
sinh |
ln[x + √(x |
(-∞, ∞) |
(-∞, ∞) |

Inverse hyperbolic cosine |
cosh |
ln[x + √(x |
[1, ∞) |
[0, ∞) |

Inverse hyperbolic tangent |
tanh |
½ ln[(1 + x)/(1 – x)] |
(-1,1) |
(-∞, ∞) |

Inverse hyperbolic cosecant |
csch |
ln[(1 + √(x |
(-∞, ∞) |
(-∞, ∞) |

Inverse hyperbolic secant |
sech |
ln[(1 + √(1 – x |
(0, 1] |
[0, ∞) |

Inverse hyperbolic cotangent |
coth |
½ ln[(x + 1)/(x – 1)] |
(-∞, -1) or (1, ∞) |
(-∞, ∞) |

**Inverse hyperbolic sine Function**

**sinh ^{-1} x = ln[x + √(x^{2} + 1)]**

**Proof:**

Let sinh

^{-1}x = z, where z ∈ R⇒ x = sinh z

Using the sine hyperbolic function we get,

⇒ x = (e

^{z}– e^{-z})/2⇒ 2x = e

^{z}– e^{-z}⇒ e

^{2z}– 2xe^{z}– 1 = 0We know that roots of an equation ax

^{2}+ bx + c = 0 are x = [-b ± √(b^{2}– 4ac)]/2aSo, e

^{z}= x ± √(x^{2 }+ 1)Since z is a real number, e must be a positive number.

Hence, e

^{z}= x + √(x^{2}+ 1)⇒ z = ln[x + √(x

^{2}+ 1)]⇒ sinh

^{-1}x = ln[x + √(x^{2}+ 1)]

sinh^{-1}x = ln[x + √(x^{2}+ 1)]

**Inverse hyperbolic cosine Function**

**cosh ^{-1} x = ln[x + √(x^{2} – 1)]**

**Proof:**

Let cosh

^{-1}x = z, where z ∈ R⇒ x = cosh z

Using the cosine hyperbolic function we get,

⇒ x = (e

^{z}+ e^{-z})/2⇒ 2x = e

^{z}+ e^{-z}⇒ e

^{2z}– 2xe^{z}+ 1 = 0We know that roots of an equation ax

^{2}+ bx + c = 0 are x = [-b ± √(b^{2}– 4ac)]/2aSo, e

^{z}= x ± √(x^{2}– 1)Since z is a real number, e must be a positive number.

Hence, e

^{z}= x + √(x^{2}– 1)⇒ z = ln[x + √(x

^{2}– 1)]⇒ cosh

^{-1}x = ln[x + √(x^{2}– 1)]

cosh^{-1}x = ln[x + √(x^{2}– 1)]

**Inverse hyperbolic tangent function**

**tanh ^{-1} x = ½ ln[(1 + x)/(1 – x)] = ½ [ln(1 + x) – ln(1 – x)]**

**Proof:**

Let tanh

^{-1}x = z, where z ∈ R⇒ x = tanh z

Using the tangent hyperbolic function we get,

tanh z = (e

^{z}– e^{-z})/(e^{z}+ e^{-z})x =

⇒ x = (e

^{2z }– 1)/(e^{2z }+ 1)⇒ x (e

^{2z}+ 1) = (e^{2z}– 1)⇒ (x – 1) e

^{2z}+ (x + 1) = 0⇒ e

^{2z}= -[(x +1)/(x – 1)]⇒ e

^{2z}= [(x + 1)/(1 – x)]⇒ 2z = ln [(x + 1)/(1 – x)]

⇒ z = ½ ln[(1 + x)/(1 – x)] = ½ [ln(1 + x) – ln(1 – x)]

⇒ tanh

^{-1}x = ½ ln[(1 + x)/(1 – x)] = ½ [ln(1 + x) – ln(1 – x)]

tanh^{-1}x = ½ ln[] = ½ [ln(1 + x) – ln(1 – x)]

**Inverse hyperbolic cosecant function**

**csch ^{-1} x = ln[(1 + √(x^{2} + 1)/x]**

**Proof:**

Let csch

^{-1}x = z, where z ∈ R⇒ x = csch z

Using the cosecant hyperbolic function we get,

csch z = 2/(e

^{z}– e-z)⇒ x = 2/(e

^{z}– e-z)⇒ x =

⇒ x = 2e

^{z}/(e^{2z}– 1)⇒ x (e

^{2z}– 1) = 2e^{z}⇒ xe

^{2z}− 2e^{z }– x = 0We know that roots of an equation ax

^{2}+ bx + c = 0 are x = [-b ± √(b^{2 }– 4ac)]/2a⇒ e

^{z}= (1 + √(x^{2}+ 1)/x⇒ z = ln[ ]

⇒ csch

^{-1}x = ln[ ] = ln[1 + √(1 + x^{2})] – ln(x)

csch^{-1}x = ln[] = ln[1 + √(1 + x^{2})] – ln(x)

**Inverse hyperbolic secant function**

**sech ^{-1} x = ln[(1 + √(1 – x^{2})/x]**

**Proof:**

Let sech

^{-1}x = z, where z ∈ R⇒ x = sech z

Using the secant hyperbolic function we get,

sech z = 2/(e

^{z}+ e^{-z})⇒ x = 2/(e

^{z}+ e^{-z})⇒ x =

⇒ x = 2e

^{z}/(e^{2z}+ 1)⇒ x (e

^{2z}+1) = 2e^{z}⇒ xe

^{2z}− 2e^{z}+ x = 0We know that roots of an equation ax

^{2 }+ bx + c = 0 are x = [-b ± √(b^{2}– 4ac)]/2aSo, by simplifying we get,

e

^{z }=z = ln[ ] = ln[1 + √(1 – x

^{2})] – ln(x)⇒ sech

^{-1}x = ln[ ] = ln[1 + √(1 – x^{2})] – ln(x)

sech^{-1}x = ln[] = ln[1 + √(1 – x^{2})] – ln(x)

**Inverse hyperbolic cotangent function**

**coth ^{-1} x = ½ ln[(x + 1)/(x – 1)]**

**Proof:**

Let coth

^{-1}x = z, where z ∈ R⇒ x = coth z

Using the cotangent hyperbolic function we get,

coth z = (e

^{z}+ e^{-z})/(e^{z }– e^{-z})⇒ x = (e

^{z}+ e-^{z})/(e^{z}– e^{-z})⇒ x =

⇒ x = (e2z + 1)/(e

^{2z }– 1)⇒ x (e

^{2z}– 1) = (e^{2z}+ 1)⇒ (x – 1) e2z – (x + 1) = 0

⇒ e

^{2z}= [(x +1)/(x – 1)]⇒ 2z = ln [(x + 1)/(x – 1)]

⇒ z = ½ ln[(x + 1)/(x – 1)] = ½[ln(x + 1) – ln(x – 1)]

coth^{-1}x = ½ ln[(x + 1)/(x – 1)] = ½ [ln(x + 1) – ln(x – 1)]

**Derivates of inverse hyperbolic functions**

Inverse hyperbolic function |
Derivative |
---|---|

sinh |
1/√(x |

cosh |
1/√(x |

tanh |
1/(1 – x |

csch |
1/{|x|√(1 + x |

sech |
-1/[x√(1 – x |

coth |
1/(1 – x |

### Sample Problems

**Problem 1: If sinh x = 4, then prove that x = log _{e}(4 + √17).**

**Solution:**

Given, sinh x = 4

⇒ x = sinh

^{-1}(4)We know that,

sinh^{-1}(x) = loge [x + √(x^{2}+ 1)]⇒ x = log

_{e}[4 + √(4^{2}+ 1)] = log_{e}(4 + √17)Hence, x = log

_{e}(4 + √17)

**Problem 2: Prove that tanh ^{-1} (sin x) = cosh^{-1} (sec x).**

**Solution:**

We know that,

tanh^{-1}x = 1/2 ln[(1+x)/(1-x)]Now, tanh-1 (sin x) = 1/2 log[(1 + sin x)/(1 – sin x)]

We have,

cosh^{-1}x = ln(x + √[x^{2}-1])Now, cosh

^{-1}(sec x) = ln[sec x + √(sec^{2}x – 1)]= ln[sec x + √tan

^{2}x]{Since, sec^{2}x – 1 = tan^{2}x}= ln[sec x + tan x]

= ln[1/cos x + sin x/cos x]

= ln[(1 + sin x)/cos x]

Now, multiply and divide the term with 2

= 1/2 × 2 ln[(1 + sin x)/cos x]

= 1/2 ln[(1 + sin x)/cos x]

^{2}{since, 2 ln x = ln x^{2}}= 1/2 ln[(1 + sin x)

^{2}/cos^{2}x]We know,

cos^{2}x = 1 – sin^{2}x = (1 + sin x)(1 – sin x)Hence, (1 + sin x)2/cos2 x = [(1 + sin x)(1+ sin x)]/[(1 + sin x)(1 – sin x)] = (1 + sin x)/(1 – sin x)

= 1/2 ln[(1 + sin x)/(1 – sin x)]

= tanh

^{-1}(sin x)Hence, tanh

^{-1}(sin x) = cosh^{-1}(sec x)

**Problem 3: Find the value of tanh ^{-1 }(1/5).**

**Solution:**

We know,

tanh^{-1 }x = 1/2 ln[(1+x)/(1-x)]⇒ tanh

^{-1 }(1/5) = 1/2 ln[(1+(1/5))/(1 – (1/5)]= 1/2 ln[(6/5)/(4/5)]

=1/2 ln(3/2)

Hence, tanh

^{-1}(1/5) = 1/2 ln(3/2)

**Problem 4: Find the value of sech ^{-1} (3/8).**

**Solution:**

We know,

sech

^{-1}x = ln[(1 + √(1 – x^{2})/x]So, sech

^{-1}(3/8) == ln[(8 + √(64 – 9))/3]

=

Hence, sech

^{-1}(3/8) =

**Problem 5: Find the derivative of [sinh ^{-1} (5x + 1)]^{2}.**

**Solution: **

Let y = [sinh

^{-1}(5x + 1)]^{2}Now derivative of the given function is,

dy/dx = d([sinh

^{-1}(5x + 1)]^{2})/dx= 2[sinh

^{-1}(5x + 1)] d/dx [sinh^{-1}(5x + 1)We know

d(sinh^{-1}x)/dx = 1/√(x^{2}+ 1)= 2 [sinh

^{-1}(5x + 1)] {1/√[(5x+1)^{2}+ 1]} d(5x+1)/dx= 2 [sinh-1 (5x + 1)] × {1/√(25×2 + 10x + 2)} × 5

= 10 sinh

^{-1}(5x+1)/[√(25x^{2}+10x+2)]Hence, the derivative of [sinh

^{-1}(5x+1)] = .