Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
[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]
[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]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We appreciate your time. Please come back anytime for the latest information and answers to your questions. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
