By the product rule, we first have
f ′ ( x ) = ( x e ( x 2 ) ) ′ = ( x ) ′ e ( x 2 ) + x ( e ( x 2 ) ) ′ = e ( x 2 ) + x ( e ( x 2 ) ) ′ . {\displaystyle f'(x)=(xe^{(x^{2})})'=(x)'e^{(x^{2})}+x(e^{(x^{2})})'=e^{(x^{2})}+x(e^{(x^{2})})'.}
Using the chain rule with u = x 2 {\displaystyle u=x^{2}} , we have
( e ( x 2 ) ) ′ = d d x e ( x 2 ) = d d u e u ⋅ d u d x = e u d ( x 2 ) d x = e ( x 2 ) ⋅ 2 x = 2 x e ( x 2 ) . {\displaystyle (e^{(x^{2})})'={\frac {d}{dx}}e^{(x^{2})}={\frac {d}{du}}e^{u}\cdot {\frac {du}{dx}}=e^{u}{\frac {d(x^{2})}{dx}}=e^{(x^{2})}\cdot 2x=2xe^{(x^{2})}.}
Therefore, plugging this back into the above equation, we get
f ′ ( x ) = e ( x 2 ) + x ( e ( x 2 ) ) ′ = e ( x 2 ) + x ⋅ ( 2 x e ( x 2 ) ) = ( 2 x 2 + 1 ) e ( x 2 ) . {\displaystyle {\begin{aligned}f'(x)=e^{(x^{2})}+x(e^{(x^{2})})'=e^{(x^{2})}+x\cdot (2xe^{(x^{2})})=\color {blue}(2x^{2}+1)e^{(x^{2})}.\end{aligned}}}
, we have
