Question

calculus help.. can someone tell me if my answer is correct??? find the derivative of the function. y=x^2e^-x

Answer 1

y(prime)=2x+e^-x*-1

my answer=2x-e^-x u=-x
u(prime)=-1

Answer 2

\[x^2 e^{-x} = \frac{x^2}{e^x}\]

Now, do the quotient rule.

Answer 3

oooooooooooooooooooooooooooooooooooo okay that makes sense...so my answer is wrong hugh?

Answer 4

oh hell no
quotient rule is a pain

Answer 5

Yeah, it seems so.

Answer 6

lololololololo

Answer 7

can i use the product rule instead?

Answer 8

\((fg)'=f'g+g'f\) just like yesterday

Answer 9

in this case \(f(x)=x^2,f'(x)=2x,g(x)=e^{-x}, g'(x)=-e^{-x}\)

Answer 10

plug and chug

Answer 11

awww u remembered! hahah okay i got it...but one question....my answer was way off right?

Answer 12

way off, right

Answer 13

:{.................okay :]

Answer 14

so my answer is this right? y(prime)=2x(e^x)-e^-x (x^2)

Answer 15

Quotient rule wasn't so hard. Just saying. \[\frac{d}{dx} \frac{f(x)}{g(x)} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}\]
It looks complicated but all you have to do is derivative factors then plug in, just like the way satellite did.

Answer 16

functions*

Answer 17

lo-dee-hi-hidee-lo-all-over-lo^2

Answer 18

got it

Answer 19

so my answer is right, right?

Answer 20

Well, my answer is \[y \prime = \frac{2x - x^2}{e^x}\]

Answer 21

same thing.....thanks!

Answer 22

Yeah your answer is right, but you can simplify it.

Answer 23

thank u!

Answer 24

No problem!