Answer
$1-4x^3$
Work Step by Step
We have: $f(x)=e^{\ln x} -e^{2 \ln (x^2)} $
We differentiate both sides with respect to $x$.
$f^{\prime}(x)=\dfrac{d}{dx} [e^{\ln x} -e^{2 \ln (x^2)}] \\=\dfrac{d}{dx} [e^{\ln x} -e^{\ln x^4} ]\\= \dfrac{d}{dx}(x-x^4)$
Simplify to obtain:
$f^{\prime}(x)=1-4x^3$
