Answer
$$ g'(x)= \frac{e^x}{1-e^{2x}}.$$
Work Step by Step
Since $ g(x)=\tanh^{-1}e^x $, then the derivative, by using the chain rule, is given by
$$ g'(x)=\frac{1}{1-(e^x)^2}(e^x)'=\frac{e^x}{1-e^{2x}}.$$
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