Answer
tanh(ln x) = $\frac{x^{2} - 1}{x^{2} +1}$
Work Step by Step
Start with the definition of $\tanh{(x)} = \frac{\sinh{(\theta)}}{\cosh{(\theta)}}$; in this case $\theta = \ln{x}$
$tanh(ln x) = \frac{sinh(ln x)}{cosh(ln x)}$
Switch for the hyperbolical sine and cosine definition:
$\sinh(\theta) = \frac{e^{\theta}-e^{-\theta}}{2}$
$\cosh{(\theta)} = \frac{e^{\theta}+e^{-\theta}}{2}$
Therefore:
$tanh(ln x) = \frac{(e^{ln x} - e^{-ln x})}{(e^{ln x} + e^{-lnx})}$
$tanh(ln x) =\frac{e^{ln x} - e^{-lnx}}{e^{lnx} + e^{-ln x}}$
Recall that:
$\theta^{\log_{\theta}\alpha} = \alpha$
$e^{\ln \theta} = \theta$
Therefore:
$tanh(ln x) =\frac{x - x^{-1}}{x + x^{-1}}$
Multiply by $\frac{x}{x}$ to eliminate the $x^{-1}$
$tanh(ln x) =\frac{x - x^{-1}}{x + x^{-1}}$ $\times$ $\frac{x}{x}$ = $\frac{x^{2} -1}{x^{2} + 1}$
