find the equation of the tangent line to the curve f(x)= lnx/e^x at x=1

Question

anonymous · Answer

slope m=f'(1)
point (1,f(1))
equation:
(y-f(1))=m(x-1)

turingtest · Answer

^right, so what have you got for f'(x) ?

liizzyliizz · Answer

wouldn't f'(x) be (1/x - ln(x)) * e^-x ?

turingtest · Answer

yeah, so what is f'(1) ?

liizzyliizz · Answer

it would be 1

liizzyliizz · Answer

?

turingtest · Answer

$$f'(x)=e^{-x}(\frac1x-\ln x)$$$$f'(1)=e^{-1}(\frac11-\ln 1)=e^{-1}(1-0)=e^{-1}$$

liizzyliizz · Answer

err I messed up on the e^-1 :/

turingtest · Answer

so that's your slope, m
 what's f(1) ?

liizzyliizz · Answer

attachment

anonymous · Answer

right, so now you have:
slope m=e^-1
and point (1,0)
replace in the equation

liizzyliizz · Answer

ok so it would be y-0=e^-1(x-1) correct?

liizzyliizz · Answer

then you would just change the form, which is what im doing now to match the choices in my question

turingtest · Answer

yeah that's right
don't know what for you need it in...

turingtest · Answer

what form*

anonymous · Answer

y=e^-1(x-1) is  the equation of the tangent line to the curve

liizzyliizz · Answer

these are the choices- x-ey-1 =0
x+ey-1=0
x-y-1=0
ex+y-1=0
ex-y-1=0

liizzyliizz · Answer

i feel like when i took this test i did the work somewhat right, but when it came to switching it out i made a careless mistake so i want to know which one was it. :c

turingtest · Answer

multiply both sides by e I meant :/

turingtest · Answer

actually something is missing...

turingtest · Answer

$$\large y=e^{-1}(x-1)$$$$\large ey=x-1$$$$\large x-ey-1=0$$but that's not a choice...

turingtest · Answer

oh, it is the first one
I didn't see that one on the first line

anonymous · Answer

liizzyliizz the first choice is -x-ey-1 =0? or  x-ey-1 =0?

liizzyliizz · Answer

first choice is x-ey-1 =0

anonymous · Answer

that is the answer

liizzyliizz · Answer

well this helped thank you, I know what I did wrong. *sigh*