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Finding the equation for the tangent line.

  1. Jul 17, 2007 #1
    1. The problem statement, all variables and given/known data
    find the equation for the tangent line.
    1. (e^x)(cos x) where x=0

    2. Relevant equations
    plugged 0 into the equation, (e^0)(cos 0) and got 1 for the y-coordinate. so I got the points(0,1). For the slope, I derived the equation into -(e^x)(sin x). then i plugged 0 in and got 0 for the slope.
    I used point-slope form, y-1=0(x-0)

    3. The attempt at a solution
    y-1=0(x-0) or y=1.
    but the answer says x-y+1=0
    I believe the slope was suppose to be 1 and I got 0, what is the right way to get the slope?
    Last edited: Jul 17, 2007
  2. jcsd
  3. Jul 17, 2007 #2


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    Homework Helper

    I doubt anyone will help you until you post a more complete problem statement.
  4. Jul 17, 2007 #3


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    Staff: Mentor

    Sorry, I'm not tracking what you are trying to do. Are you saying that you have the following equation:

    [tex]y(x) = {e^x} cos(x)[/tex]

    and you want to know what the tangent (derivative) is at x=0? Do you know how to take the derivative of that y(x) function? Also, you won't be plugging in x=0 until you have that final derivative function....
  5. Jul 17, 2007 #4
    I saw my mistake, I did the derivative wrong. sorry
  6. Jul 17, 2007 #5


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    Staff: Mentor

    No need to be sorry. So that means you're okay now?
  7. Jul 17, 2007 #6
    For this problem, Yes
  8. Jul 17, 2007 #7
    Yeah, you forgot the chain rule.
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