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AP Calc. question

  1. Nov 22, 2007 #1
    1. The problem statement, all variables and given/known data

    If xe^y + ycosx = 1 defines y as a function of x, find y(0)

    3. The attempt at a solution

    My problem is getting it into y= form. Since there is an e^y, I thought about taking a natural log, but that doesn't seem to be getting me anywhere. Help?
  2. jcsd
  3. Nov 22, 2007 #2


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    You aren't asked to solve for y in general. When x= 0, you have 0*e^y+ y cos(0)= 1. Solve that for y.
  4. Nov 22, 2007 #3
    Thank you! That makes so much more sense. I just had a big Duh! moment right now. :blushing:
  5. Nov 22, 2007 #4
    Okay, so I have two more parts to the problem: find y'(0) and y''(0)

    For y'(0), I took the derivative and got the formula y'= (e^y - ysinx)/(-xe^y - cosx) I plugged in zero for x and 1 for y, and got e as the answer. I'm not sure that this is correct.

    For y''(0), I painstakingly took the derivative again and plugged in zero and 1 as before and got 1, which our math teacher said was answer we should definitely not get. I'm not sure my methods are currect. Can someone help me?
    Last edited: Nov 22, 2007
  6. Nov 23, 2007 #5
    When I did it I got [tex] \frac{dy}{dx}(0) = -e [/tex] and [tex] \frac{d^2 y}{dx^2} (0) = 2e^2+1 [/tex]

    Edit: You're equation for y' is correct, so you should've gotten -e as well
    Last edited: Nov 23, 2007
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