Question

Differentiate y = xe^(-kx)

Answer 1

\[\bf y'=e^{-kx}-kxe^{-kx}\]@suckmylee

Answer 2

@suckmylee

Answer 3

How did you get that -k before the e^-kx

Answer 4

chain rule

Answer 5

\[\frac{d}{dx}[e^{f(x)}]=f'(x)e^{f(x)}\] by the chain rule

Answer 6

OH
e^-kx * d/dx -kx
right?

Answer 7

and the derivative of \(-kx\) is \(-k\)

Answer 8

yes

Answer 9

THANK YOU

Answer 10

yw

Answer 11

the answer is e^(-kx)(-kx+1) I'm a little bit confused where the 1 came from

Answer 12

do you see how to get
\[ y'=e^{-kx}-kxe^{-kx} \]
using the product rule and the chain rule ?
notice you can factor e^(-kx) from both terms

\[ y'=e^{-kx}(1-kx) \]

and you can distribute the e^(-kx) to get back the other form.