Question

Determine the equation of th tangent line to the function f(x)=x/e^2x at the point (1,1/e^2)

Answer 1

take the derivative of f(x) and plug 1 into your derivative to find the slope of your tangent line at that point

Answer 2

but how do I derive (x/e^2x)?? thats where I am stuck

Answer 3

wait isnt it using the quotient rule so the denominator should be e^2x squared right??

Answer 4

do you know how to differentiate
\[ f(x)=\frac{x}{e^{2x}} \]

Answer 5

quotient rule right ??

Answer 6

1* e^2x+x*2e^2x/e^2x squared??

Answer 7

yep ... now put, x=1, what do you get?

Answer 8

i mean put x=1 on your derivative.

Answer 9

okay i need help beacuase i would be guessing
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Answer 10

Dang!! just replace x=1 on
1* e^2x+x*2e^2x/e^2x squared??

Answer 11

i get e^2 =2e^2/(e^2)^2

Answer 12

can you help me experiment,please?

Answer 13

?? no ...

Answer 14

you made some mistake somewhere
wolf says this is zero.
http://www.wolframalpha.com/input/?i=D%5Bx%2Fe%5Ex%2C+x%5D%2F.x-%3E1

Answer 15

also you messed up your derivative somewhere
http://www.wolframalpha.com/input/?i=D%5Bx%2Fe%5Ex%2C+x%5D

Answer 16

i give up

Answer 17

i got -1/e^2 after plugging 1 into the derivative

Answer 18

travis can you do it step by step if its not too much.

Answer 19

do your derivative again ... wolf says the derivative is
\[ \frac{1}{e^x} - \frac{x}{e^{x}}\]
you probably messed up somewhere.

Answer 20

Woops!! my fault ... i missed e^2x
http://www.wolframalpha.com/input/?i=D%5Bx%2Fe%5E%282x%29%2C+x%5D%2F.x-%3E1

Answer 21

now use this formula
\[ y - y' = f'(x') (x-x')\]
replace \((x', y') \to (1, 1/e) \) and replace your \( f'(1) \) by above.

Answer 22

i used the quotient rule
f'(x)=(e^2x-x2e^2x)/(e^2x)^2

Answer 23

plug in 1 and you get (e^2-2e^2)/e^4

Answer 24

this simplifies to (-e^2)/(e^4)
which simplifies to -1/e^2