Question

what is the first and second derivative of xe^(2x)?

Answer 1

Remember the product rule?
(fg)'=f'g+fg'
You have to use it.
Try with f=x and g=e^(2x)
Do you know how to go on?

Answer 2

not really, so it's not 2xe^2x?

Answer 3

No, you have to use the product rule.
Find this first:
f=x ---> f'=?
g=e^(2x) ---> g'=?
What did you get?

Answer 4

x^2 and 2e^2x

Answer 5

x^(2)e^(2x) + 2xe^(2x)?

Answer 6

f=x -------> f'=1
g=e^(2x)---->g'=2e^(2x)
Plug that on this formula:
f'g+fg'

Answer 7

e^2x + 2xe^2x

Answer 8

so the next derivative would be 2e^2x + 4xe^2x?

Answer 9

\[\large xe^{2x}\]
\[\Large y' = 2xe^{2x} + e^{2x}\]
\[\LARGE y'' = 2(xe^{2x} + e^{2x}) + e^{2x} \implies 2xe^{2x} + 2e^{2x} +e^{2x}\]
if my solution is right...that means your answer is wrong..

Answer 10

i don't understand how you can pull the 2 out like that

Answer 11

constant multiple rule...

Answer 12

are you familiar with that rule?

Answer 13

no

Answer 14

\[\huge \frac{d}{dx} (ax) \implies a [\frac{d}{dx} (x)]\]
where a is a constant

Answer 15

for example derivative of 2x^2 is

2( derivative of x^2)

derivative of x^2 = 2x so

2(derivative of x^2) = 2(2x) = 4x

Answer 16

i think you need to review the rules of derivatives http://www.math.brown.edu/UTRA/derivrules.html

Answer 17

how the helll am i supposed to plug that into a Wronskian?!

Answer 18

hmm wait...i missed something..
\[\Large 2(2xe^{2x} + e^{2x}) + e^{2x} \implies 4xe^{2x} + 2e^{2x} + e^{2x}\]

Answer 19

now combine.. \[\large \implies 4xe^{2x} + 3e^{2x}\]
now do your wronskian thingy

Answer 20

this is bull.... when am i really going to use the wronksian and cramer when I am an actuarry?!

Answer 21

this one problem is taking forever!