Deriving the formula for arctanh(x)

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SUMMARY

The discussion focuses on deriving the formula for the inverse hyperbolic tangent, arctanh(x), starting from the definition of the hyperbolic tangent function, tanh(x), expressed in terms of exponentials. The equation to prove is arctanh(x) = 1/2 log((1+x)/(1-x)). The user seeks assistance in proving this relationship, indicating a need for clarity in the steps involved in manipulating the exponential form of tanh(x) to isolate x.

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  • Understanding of hyperbolic functions, specifically tanh(x)
  • Familiarity with exponential functions and their properties
  • Knowledge of logarithmic identities and their applications
  • Basic skills in algebraic manipulation and solving equations
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  • Study the properties of hyperbolic functions, particularly tanh(x) and its inverse
  • Learn about the derivation of inverse functions in calculus
  • Explore logarithmic identities and their proofs
  • Practice solving equations involving exponential and logarithmic forms
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Students studying calculus, mathematics enthusiasts, and anyone interested in understanding hyperbolic functions and their inverses.

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Homework Statement



"starting from the definition of tanhx in terms of exponentials, prove that...

arctanh(x)=1/2log((1+x)/(1-x))

This is such a simple equation but I'm having a hard time proving it, can anyone help?

Cheers


Homework Equations





The Attempt at a Solution

 
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If y= tanh(x) then
y= \frac{e^x- e^{-x}}{e^x+ e^{-x}}
Now use the standard method of finding an inverse function: solve that equation for x.
 

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