Question

I do not undestand !!!!! Sniff sniff find the derivative of the function.
y=x^2*e^x-2e^x+2e^x

Answer 1

lorda mercy

Answer 2

:]

Answer 3

and this JUST HW :[

Answer 4

first one requires the product rule, second two do not

Answer 5

(crying)........lol

Answer 6

hold the phone
does it really say
\[y=x^2e^x-2e^x+2e^x\]

Answer 7

because \(-2e^x+2e^x=0\) is why i am asking, so this is kind of dopey

Answer 8

so soory its -2xe^x

Answer 9

\[y=x^2e^x-2xe^x+2e^x\]makes more sense, yes

Answer 10

sorry

Answer 11

step by step (inch by inch) you need the product rule
\[(fg)'=f'g+g'f\] for the first two terms

Answer 12

first term is
\[x^2e^x\] use \((fg)'=f'g+g'f\) with \(f(x)=x^2,f'(x)=2x,g(x)=e^x, g'(x)=e^x\)

Answer 13

umm in the worked out solution thing 4 this book it says to take out e^x??

Answer 14

you will have an \(e^x\) in each term, yes, so you can factor it out
that part is algebra

Answer 15

lets get the answer written first, do the algebra second

Answer 16

\[y=x^2 \times e^x-2xe^x+2e^x \]
use chain rule in both,
it will be
\[(x^2e^x + e^x \times 2x) - (2xe^x + e^x \times 2) + 2e 6x\]

Answer 17

so i could do that as the first step then?

Answer 18

i wouldn't, no
i would do the algebra last
no need to introduce another product
of course you will get the same answer

Answer 19

okay then

Answer 20

i would just go ahead and grind it til i find it

Answer 21

lol thats funny

Answer 22

first term give
\[2xe^x+x^2e^x\] in any order you choose
(drivers ed instructor used that line when driving a stick shift)

Answer 23

okay

Answer 24

can we stick to orginal plz i feel i'll get confused :[

Answer 25

or if you take out e^x
you get
\[y=e^x(x^2 -2x+2)\]
the differentiation will be
\[e^x(2x -2) + (x^2 - 2x - 2)e^x\]

Answer 26

second one, again a product, so again the product rule
\[-2(xe^x+e^x)\]

Answer 27

and third term is just \(2e^x\)

Answer 28

okay so ur using the second part of the function?

Answer 29

don't forget that \(-2\) for the whole product

Answer 30

first or second term?

Answer 31

Satellite 73 ur losing me!

Answer 32

ok lets go sloooooooow

Answer 33

k :]

Answer 34

first term is \(x^2e^x\) right? we tackle that one first

Answer 35

k

Answer 36

this is a product, so we need the product rule

Answer 37

okay

Answer 38

the product is \(x^2\) times \(e^x\) so in english, we take the derivative of one of them, multiply it by the other (not the derivative, just the other function) and then repeat the process in the reverse order

Answer 39

it'll be 2x(e^x)+E^x(x^2) right?

Answer 40

yes

Answer 41

yeah!

Answer 42

whew
now on to \(-2xe^x\)

Answer 43

Im tiring u out sorry

Answer 44

pull that \(-2\) right out front
not at all

Answer 45

so again we have a product, this time it is \(x\) times \(e^x\) so we repeat the process as before
let me know what you get

Answer 46

okay so let me get this straight from the product of the first term we have 2x(e^x)+e^x(x^2) right?

Answer 47

yes

Answer 48

and then we take care of -2xe^2 right?

Answer 49

generally it looks nicer to write \(2xe^x+x^2e^x\) but no matter, multiplication is commutative

Answer 50

yes, that is next

Answer 51

okay sorry so we apply the product rule to -2xe^x so it'l be -2(e^x)+e^x(-2x) right?

Answer 52

yes yes yes

Answer 53

i mean -2(e^x)

Answer 54

\[-2xe^x-2e^x\] or any combination of those

Answer 55

its -2xe^x+2E^x isn't it?

Answer 56

both minus by the distributive law

Answer 57

u lost me :[

Answer 58

how? or why?

Answer 59

\[-2xe^x\] is what you have
each part in the product rule will have a \(-2\) in it

Answer 60

\[-2(xe^x)\] derivative is \(-2(xe^x+e^x)=-2xe^x-2e^x\) by the distributive law

Answer 61

u mean taking out a -2 right? from xe^x+e^x right?

Answer 62

yes

Answer 63

YES! okay i get u now>

Answer 64

so from there do i treat each to the product rule?

Answer 65

not the last term , because it is not a product of two functions, it is just \(2e^x\) so the derivative will also be \(2e^x\)

Answer 66

ok so i treat the second to the last term to the product rule right?

Answer 67

we have taken care of \(x^2e^x\) and got
\[x^2e^x+2xe^x\] in whatever order we write it
then we took care of \(-2xe^x\) and got \(-2xe^x-2e^x\)

Answer 68

yes, but we did that one already

Answer 69

don't do it twice!!

Answer 70

ok so from here this is all?

Answer 71

yes, put it all together

Answer 72

AHHHHHHHHHHHHHHHHHHHHHHHH! Thanks!

Answer 73

\[x^2e^x+2xe^x-2xe^x-2e^x+2e^x\] probably some algebra will clean this up a great deal and you will end up with very little

Answer 74

okay thanks so much

Answer 75

yw
in fact, it looks like you only end up with \(x^2e^x\) hmmm it must be late

Answer 76

yeah thanks

Answer 77

not reaallly its only 9:17 pm here in Cali. how late is it where ur from?

Answer 78

do you know the way to san jose?

Answer 79

it is 12:20 or so, i am out