# Stationary points of y=-sinx+cosx

by pip_beard
Tags: cosx, points, stationary, ysinx
 P: 14 1. The problem statement, all variables and given/known data Solve the stationary points of y=-sinx+cosx for domain -pi
 P: 104 cosx+sinx=0 cosx=-sinx 1=-tanx??
 P: 14 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???
Mentor
P: 21,314
Stationary points of y=-sinx+cosx

 Quote by pip_beard 1. The problem statement, all variables and given/known data Solve the stationary points of y=-sinx+cosx for domain -pi
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.
Mentor
P: 21,314
 Quote by pip_beard because surly the x co-ordinate is 0???
Why would you think this?
 P: 14 so therefore: dy/dx=cosx+sinx. stationary points when diff = 0 so cosx+sinx=0 where do i go from here??
P: 14
 Quote by Mark44 Why would you think this?
because stationary points lie on the x axis??
Mentor
P: 21,314
 Quote by pip_beard because stationary points lie on the x axis??
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.
Mentor
P: 21,314
 Quote by pip_beard so therefore: dy/dx=cosx+sinx.
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?

 Quote by pip_beard stationary points when diff = 0 so cosx+sinx=0 where do i go from here??

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