## Calculus 8th Edition

$\tanh(ln x) = \dfrac{x^{2} - 1}{x^{2} +1}$
Since, we know $\tanh{(\theta)} = \dfrac{\sinh{(\theta)}}{\cosh{(\theta)}}$ Here, $\theta = \ln{x}$ Thus, $\tanh(ln x) = \dfrac{sinh(ln x)}{cosh(ln x)} = \dfrac{(e^{ln x} - e^{-ln x})}{(e^{ln x} + e^{-lnx})} =\dfrac{e^{ln x} - e^{-lnx}}{e^{lnx} + e^{-ln x}}$ Also, $\theta^{\log_{\theta}\alpha} = \alpha$ so, $e^{\ln \theta} = \theta$ Now, $\tanh(ln x) =\dfrac{x - x^{-1}}{x + x^{-1}}$ or, $=(\dfrac{x - x^{-1}}{x + x^{-1}})(\dfrac{x}{x})$ or, $=\dfrac{x^{2} -1}{x^{2} + 1}$