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