Answer
$2(x-1)$
Work Step by Step
We have: $f(x)=[\ln(e^x)]^2 -[\ln (e^x)^2] $
We differentiate both sides with respect to $x$.
$f^{\prime}(x)=\dfrac{d}{dx} [[\ln(e^x)]^2 -[\ln (e^x)^2]]\\=\dfrac{d}{dx} [[\ln(e^x)]^2 -[\ln (e^{2x})]]\\= \dfrac{d}{dx}(x^2-2x)\\=2x-2$
Simplify to obtain:
$f^{\prime}(x)=2(x-1)$
