Question

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

Answer 1

differentiate the function u have.... can u do that ?

Answer 2

im stuck on that part help

Answer 3

do u use quotient
rule

Answer 4

\[\frac{ d f(x) }{ dx } = \frac{ e^x \frac{ d \ln x }{ dx } - lnx \frac{ d e^x }{ dx }}{ (e^x)^{2} }\]

Answer 5

\[\frac{ d f(x) }{ dx } = \frac{ e^x \frac{ 1 }{ x } -( lnx \times e^x ) }{ (e^x)^{2} }\]

Answer 6

f'(x)=\[f'\left( x \right)=\frac{ e ^{x}\frac{ 1 }{ x }-\ln x e ^{x} }{ e ^{2x} }\]

Answer 7

ok thats what i got now what ??

Answer 8

lost????

Answer 9

\[\frac{ d f(x) }{dx } = \frac{ e^x -(x \times lnx \times e^x ) }{ x (e^x)^{2} }\]

Answer 10

now substitute x= 1 in that equation

Answer 11

kk

Answer 12

isnt it (e^x)^2 for the bottom ?

Answer 13

\[\frac{ d f(x) }{ dx} = \frac{ e^1 - ( 1 \times \ln 1 \times e^1) }{1 \times (e^1)^{2}}\]

Answer 14

i just make a common denominator by expanding
e^x * (1/x) - ln x *e^x

Answer 15

e^1 = e
ln 1 = 0

Answer 16

when x=1,
\[f'(1)=\frac{ e }{ e ^{2} }=\frac{ 1 }{ e }\]
this is the slope at x=1
y=f(x)=ln1/e^1=0/e=0
point is (1,0)

Answer 17

which means
\[\frac{ df(x) }{ dx } = \frac{ e }{ e^2 } = \frac{ 1 }{ e}\]

Answer 18

haha... u 've got help from 2 people at once... so i guess u have no doubts... XD

Answer 19

yea i di thank you soo 1/e is the slope??/

Answer 20

u r welcome... and yes

Answer 21

thank you isuri <3

Answer 22

hey.. im isuru... and isuri is a name for girls in our country :/

Answer 23

awesoem where are you from ??

Answer 24

SL ...

Answer 25

What lol

Answer 26

u know the asian country ...SRI LANKA

Answer 27

never heard of it lol. I live in the U.S thats why ur teaching me lol

Answer 28

XD ... but it sad to hear that u have never herd of it .:(

Answer 29

well i Google it lol. its small lol

Answer 30

it's a island... and a small one... but it's beautifull

Answer 31

yea it should be lol. its like 10 degrees feranheit where im from

Answer 32

Figure this one out alexis? c:

Answer 33

good i am surjit and i am from punjab india living in USA

Answer 34

actually the final answer asks me for the equation of a tangent line. and my answer and multiple choice don't match ):

Answer 35

uh ohs! :U

Answer 36

wt a the choices ?

Answer 37

So you found the `slope` of your tangent line.

Are the choices given in point-slope form `y-yo = m(x-xo)`
or in slope-intercept form `y=mx+b`

Answer 38

y-0=1/e(x-1)
y=1/e(x-1)

Answer 39

(a) x-ey-1=0
b x+ey-y=0
c x-y-1=0
d ex+y-1=0
e ex-y-1=0

Answer 40

ey=x-1
orx-ey-1=0

Answer 41

ok suri than you <3 im done bothering people lol

Answer 42

a is the answer.

Answer 43

thanks your the best

Answer 44

yw