tanh (log x) = | vedclass

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(C) We know that $	anh(y) = \frac{e^y - e^{-y}}{e^y + e^{-y}}$.Substituting $y = \log x$,we get:$	anh(\log x) = \frac{e^{\log x} - e^{-\log x}}{e^{\

Vedclass · Accepted Answer

$\frac{x^2-1}{x^2+1}$

Vedclass · Answer

(C) We know that $	anh(y) = \frac{e^y - e^{-y}}{e^y + e^{-y}}$.Substituting $y = \log x$,we get:$	anh(\log x) = \frac{e^{\log x} - e^{-\log x}}{e^{\log x} + e^{-\log x}}$Since $e^{\log x} = x$ and $e^{-\log x} = e^{\log(x^{-1})} = \frac{1}{x}$,the expression becomes:$	anh(\log x) = \frac{x - \frac{1}{x}}{x + \frac{1}{x}}$Multiplying the numerator and denominator by $x$,we get:$	anh(\log x) = \frac{x^2 - 1}{x^2 + 1}$

$\tanh (\log x) = $

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