Question

Find dy/dx if y= (1+x)e^x^2

Answer 1

$y=(1+x)e^{x^2}\\ y'=(1+x)'\cdot e^{x^2}+(1+x)\cdot(e^{x^2})'\\ y'=1\cdot e^{x^2}+(1+x)\cdot e^{x^2}\cdot (x^2)'\\ y'=e^{x^2}+(1+x)e^{x^2}\cdot2x\\ y'=e^{x^2}(1+(1+x)\cdot2x)\\ y'=e^{x^2}(1+2x+2x^2)\\ y'=e^{x^2}(2x^2+2x+1)$

Answer 2

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:

$\frac { dy }{ dx } =\left( 1+x \right) \cdot 2x{ e }^{ { x }^{ 2 } }+{ e }^{ { x }^{ 2 } }\cdot 1\\ \\ =2x{ e }^{ { x }^{ 2 } }\left( 1+x \right) +{ e }^{ { x }^{ 2 } }$

$\\ \\ ={ e }^{ { x }^{ 2 } }\left( 2x\left( 1+x \right) +1 \right) \\ \\ ={ e }^{ { x }^{ 2 } }\left( 2x+2{ x }^{ 2 }+1 \right) \\ \\ ={ e }^{ { x }^{ 2 } }\left( 2{ x }^{ 2 }+2x+1 \right)$