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Illusion [34]
3 years ago
13

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

Mathematics
2 answers:
nasty-shy [4]3 years ago
5 0
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)\\
Andrei [34K]3 years ago
3 0
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)
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