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

Sagot :

konrad509
[tex]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)\\[/tex]
You first need to know that:

[tex]If\quad y=u\cdot v\\ \\ \frac { dy }{ dx } =u\frac { dv }{ dx } +v\frac { du }{ dx } \\ \\ [/tex]

Knowing that u is a function of x and that v is a function of x.

So:

[tex]y=\left( 1+x \right) { e }^{ { x }^{ 2 } }=u\cdot v\\ \\ u=1+x,\\ \\ \therefore \quad \frac { du }{ dx } =1[/tex]

[tex]\\ \\ 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[/tex]

[tex]\\ \\ \therefore \quad \frac { dv }{ dp } \cdot \frac { dp }{ dx } =2x{ e }^{ { x }^{ 2 } }=\frac { dv }{ dx }[/tex]

And this means that:

[tex]\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 } }[/tex]

[tex]\\ \\ ={ 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) [/tex]
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