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MathidiaMathematics Academy
Hyperbolic functions
Lesson 33 of 75
Standard practice

Inverse hyperbolic functions

Objective. Express inverse forms logarithmically.

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  2. 2. Worked example
  3. 3. Practice
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Learn: the key idea

The inverse hyperbolic functions, such as arsinh x and arcosh x, can be expressed as natural logarithms by solving the defining exponential equation for x. For y = sinh⁻¹x, setting x = (eʸ−e⁻ʸ)/2 and solving the resulting quadratic in eʸ gives arsinh x = ln(x + √(x²+1)). Similarly arcosh x = ln(x + √(x²−1)) for x ≥ 1. These logarithmic forms allow inverse hyperbolic expressions to be evaluated, differentiated and manipulated algebraically using logarithm rules.

Worked example

Show that arsinh x = ln(x + √(x²+1)) by solving x = sinh y for y.

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Illustrated card showing 1
  1. 1Start with x = (eʸ − e⁻ʸ)/2, so 2x = eʸ − e⁻ʸ
  2. 2Multiply by eʸ: 2x eʸ = e^(2y) − 1, giving e^(2y) − 2x eʸ − 1 = 0
  3. 3Treat as a quadratic in eʸ: eʸ = [2x ± √(4x²+4)]/2 = x ± √(x²+1)
  4. 4Since eʸ>0, take the positive root: eʸ = x + √(x²+1), so y = ln(x + √(x²+1))

Answer: arsinh x = ln(x + √(x²+1)), proved

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Common mistake. Students write arcosh x = ln(x + √(x²+1)), confusing it with the formula for arsinh x. The sign inside the square root differs: arsinh uses +1, arcosh uses −1, and must be kept distinct.

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