# Stationary points of y=-sinx+cosx

1. Feb 8, 2010

### pip_beard

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. Feb 8, 2010

cosx+sinx=0
cosx=-sinx
1=-tanx??

3. Feb 8, 2010

### pip_beard

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???

4. Feb 8, 2010

### 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.

5. Feb 8, 2010

### Staff: Mentor

Why would you think this?

6. Feb 8, 2010

### pip_beard

so therefore:

dy/dx=cosx+sinx.

stationary points when diff = 0

so cosx+sinx=0 where do i go from here??

7. Feb 8, 2010

### pip_beard

because stationary points lie on the x axis??

8. Feb 8, 2010

### 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.

9. Feb 8, 2010

### 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?