Question

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

Answer 1

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)`

Answer 2

`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)\\`