You first need to know that:
\(If\quad y=u\cdot v\\ \\ \frac { dy }{ dx } =u\frac { dv }{ dx } +v\frac { du }{ dx } \\ \\\)
Knowing that u is a function of x and that v is a function of x.
So:
\(y=\left( 1+x \right) { e }^{ { x }^{ 2 } }=u\cdot v\\ \\ u=1+x,\\ \\ \therefore \quad \frac { du }{ dx } =1\)
\(\\ \\ v={ e }^{ { x }^{ 2 } }={ e }^{ p }\\ \\ \therefore \quad \frac { dv }{ dp } ={ e }^{ p }={ e }^{ { x }^{ 2 } }\\ \\ p={ x }^{ 2 }\\ \\ \\ \therefore \quad \frac { dp }{ dx } =2x\)
\(\\ \\ \therefore \quad \frac { dv }{ dp } \cdot \frac { dp }{ dx } =2x{ e }^{ { x }^{ 2 } }=\frac { dv }{ dx }\)
And this means that:
\(\frac { dy }{ dx } =\left( 1+x \right) \cdot 2x{ e }^{ { x }^{ 2 } }+{ e }^{ { x }^{ 2 } }\cdot 1\\ \\ =2x{ e }^{ { x }^{ 2 } }\left( 1+x \right) +{ e }^{ { x }^{ 2 } }\)
\(\\ \\ ={ e }^{ { x }^{ 2 } }\left( 2x\left( 1+x \right) +1 \right) \\ \\ ={ e }^{ { x }^{ 2 } }\left( 2x+2{ x }^{ 2 }+1 \right) \\ \\ ={ e }^{ { x }^{ 2 } }\left( 2{ x }^{ 2 }+2x+1 \right)\)
