Explain how to find this derivative? (1+xe^(2x))'?

(1+xe^(2x))' = (e^(2x) + 2xe^(2x))
or
http://www.wolframalpha.com/input/?i=%28...

I'm confused as to how do you get to it? What rule are you using? And why doesn't the 1 just become 0? Please explain the steps.

f(x) = 1 + x.e^(2x)

f(x) = 1 + [x * e^(2x)]

f(x) = 1 + [x * e^(x + x)]

f(x) = 1 + [x * e^(x) * e^(x)]


The derivative of 1 is 0


Let's calculate the derivative of the expression [x * e^(x) * e^(x)] → { x * e^(x) } * e^(x)

This expression looks like (u.v), so its derivative looks like (u'.v) + (v'.u) → where:

u = x * e^(x)

v = e^(x) → v' = e^(x)

[x * e^(x) * e^(x)]' = (u'.v) + (v'.u)

[x * e^(x) * e^(x)]' = { u' * e^(x) } + { e^(x) * x * e^(x) } ← memorize this result as (1)



Let's calculate the derivative of the expression u = x * e^(x)

This expression looks like (a.b), so its derivative looks like (a'.b) + (b'.a) → where:

a = x → a' = 1

b = e^(x) → b' = e^(x)


u' = (a'.b) + (b'.a)

u' = [1 * e^(x)] + [e^(x) * x]

u' = e^(x) + [e^(x) * x]

u' = e^(x) * (1 + x) ← memorize this result as (2)



Restart from (1)

[x * e^(x) * e^(x)]' = { u' * e^(x) } + { e^(x) * x * e^(x) } → recall (2)

[x * e^(x) * e^(x)]' = { e^(x) * (1 + x) * e^(x) } + { e^(x) * x * e^(x) }

[x * e^(x) * e^(x)]' = { e^(x) * e^(x) * (1 + x) } + { e^(x) * e^(x) * x }

[x * e^(x) * e^(x)]' = [e^(x) * e^(x)] * [(1 + x) + x]

[x * e^(x) * e^(x)]' = [e^(x) * e^(x)] * [1 + x + x]

[x * e^(x) * e^(x)]' = [e^(x) * e^(x)] * (1 + 2x)

[x * e^(x) * e^(x)]' = [e^(x + x)] * (1 + 2x)

[x * e^(x) * e^(x)]' = e^(2x) * (1 + 2x)

[x * e^(x) * e^(x)]' = e^(2x) + 2x.e^(2x)



Conclusion:

f'(x) = e^(2x) + 2x.e^(2x)

f'(x) = e^(2x) * (1 + 2x) ← this form is more convenient to study the sign