Question

when I find the deriative of (x-2)e^-x I am using the product rule and get -1. The answer is 1. I am missing a negative somewhere. Can you please help

Answer 1

Show your work?

Answer 2

(x-2)(-e^-x)+e^-x (1)

Answer 3

mots likely you forgot to apply the chain rule on e^-x

Answer 4

(d/dx)(e^-x) = -e^-x

Answer 5

yeah
\[\frac{ d }{ dx }e^{-ax}=-ae^{-ax}\]

Answer 6

got that?

Answer 7

got that but I am still getting -1. I applied the product rule and the chain rule for e^-x.
Work
[f'(x)= (x+2)(-e ^{-x})chain rule + e ^{-x}(1)] Product Rule
[f'(x)=(-xe^-x)-(2e^-x)+(e^-x)]
[f'(x)=(-xe^-x)-(e^-x)]
I forgot to tell you I need to substitute 0 for x
f'(0)=(-0e^-0)-(e^-0)
f'(0)=(0) -(1) = -1
the answer is 1
please let me know how I missed a negative

Answer 8

\[f(x)=(x-2)e^{-x}\]
\[f'(x)=(x-2)[\frac{ d }{ dx}e^{-x}]+e^{-x}*\frac{ d }{ dx }(x-2)\]
\[f'(x)=(x-2)[-e^{-x}]+e^{-x}=-e^{-x}[x-3]\]
ok so far?

Answer 9

now if u put x=0 then
\[f'(0)=-e^{0}[0-3]=3\]
thats ur answer it is not (1)

Answer 10

Thank you. You showed me how my Algebra was off. The problem was f(x)=(x+2)e^-x x=0. You provided me f(x)=(x-2)e^-x but I was still easily able to follow and got my answer of 1. Thanks again