Stationary points of y=-sinx+cosx

  1. Feb 8, 2010 #1
    1. The problem statement, all variables and given/known data
    Solve the stationary points of y=-sinx+cosx for domain -pi<x<pi


    2. Relevant equations



    3. The attempt at a solution
    Differentiate: d/dx=cosx+sinx But how do i solve?
     
  2. jcsd
  3. Feb 8, 2010 #2
    cosx+sinx=0
    cosx=-sinx
    1=-tanx??
     
  4. Feb 8, 2010 #3
    ive got the answers as.. (-pi/4,-rt2) and (3pi/4, rt2) Are the answers wrong?

    I dont know how they got these??

    because surly the x co-ordinate is 0???
     
  5. Feb 8, 2010 #4

    Mark44

    Staff: Mentor

    If y = -sinx + cosx, what is dy/dx? Note that it is incorrect to say "d/dx = ..."

    If you meant dy/dx = ..., you have made a mistake. Try again.

    Also, your notation is not correct. d/dx is an operator that is written to the left of some function. In contrast, dy/dx is the derivative of y with respect to x.
     
  6. Feb 8, 2010 #5

    Mark44

    Staff: Mentor

    Why would you think this?
     
  7. Feb 8, 2010 #6
    so therefore:

    dy/dx=cosx+sinx.

    stationary points when diff = 0

    so cosx+sinx=0 where do i go from here??
     
  8. Feb 8, 2010 #7
    because stationary points lie on the x axis??
     
  9. Feb 8, 2010 #8

    Mark44

    Staff: Mentor

    x-values lie on the x-axis, but stationary points lie on the curve, which might not even touch the x-axis. For example, the only stationary point on the graph of y = x^2 + 1 is at (0, 1). This is not a point on the x-axis.
     
  10. Feb 8, 2010 #9

    Mark44

    Staff: Mentor

    No, dy/dx = -cosx - sinx

    To find the stationary points, set dy/dx to zero.
    -cosx - sinx = 0
    ==> cosx + sinx = 0
    ==> 1 + tanx = 0 (dividing both sides by cosx)
    Can you continue?

     
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