Question

derivative of y=xe^-x

Answer 1

do you know the product rule ?

Answer 2

yeah, so x would be u
and e^-x would be v?

Answer 3

yes

Answer 4

what would the deriv of e^-x be though?

Answer 5

\[ \frac{d}{dx} e^u= e^u \frac{d}{dx}u\]

Answer 6

i'm confused?

Answer 7

the derivative of e^-x would be -e^-x

Answer 8

d/dx e^x is e^x (this is a very special property)

but here are the details:
\[ \frac{d}{dx}e^x = e^x \frac{d}{dx}x= e^x \cdot 1 = e^x \]

if you have
\[ \frac{d}{dx}e^{2x} = e^{2x} \frac{d}{dx}2x= e^{2x} \cdot 2 = 2e^{2x} \]

Answer 9

so always copy the e^stuff, and then multiply by the derivative of "stuff"

Answer 10

phi can you help me on the Newton's method please

Answer 11

so the deriv of e^-x is -e^-x?

Answer 12

yes

Answer 13

so then i have..
x(-e^-x) + e^-x ..?

Answer 14

d e^(-x) is copy the e^-x then multiply by d/dx (-x) = -1 to get -1 * e^-x

Answer 15

yes that looks good. I would write it
\[ -x e^{-x}+e^{-x} \]
or
\[ e^{-x}(1-x) \]
or
\[ \frac{1-x}{e^x}\]

Answer 16

got it. thanks!