# Homework Help: Sketch the graph of y = sinx(1-cosx)

1. Jul 11, 2011

### Glissando

1. The problem statement, all variables and given/known data
Sketch the graph of y = sinx(1-cosx) over [-2pi, 2pi]

2. Relevant equations
first derivative and second derivative

3. The attempt at a solution

y' = cosxsinx
0 = cosxsinx
x = pi/2, 3pi/2, 0, pi, 2pi

I know I have the sin values right (the answers say the same) but the cos values in the answers are actually 2pi/3 for MAX and 4pi/3 for MIN. For that cosx would have to equal -1/2. I'm not sure how to get that );

Thank you!

2. Jul 11, 2011

### micromass

Hi Glissando!

Your derivative is wrong. If you want to differentiate a product, then you can't just differentiate both factors and be done with it. You need to apply the product rule:

$$(fg)^\prime=f^\prime g+fg^\prime$$

3. Jul 11, 2011

### SammyS

Staff Emeritus
Once you get the first derivative right, don't forget to find the second derivative.

4. Jul 11, 2011

### Glissando

Good point (wow that was embarassing...)

K well I tried doing it and I still get cosx = 0:

y' = cosx(1-cosx) + sin2x
y' = cosx - cos2x + sin2x
0 = -cosx + cos2x - sin2x
0 = -cosx + cos2x
0 = cosx, x = pi/2, 3pi/2

UGH ): Thanks guys for putting up with me!

5. Jul 11, 2011

### SammyS

Staff Emeritus
Well, you have y' right, but ...

-cos(x) + cos(2x) ≠ cos(x)

(Plug x = π/2 into your equation for y' to check your answer.)

You could use sin2(x) =1 - cos2(x), then y' = 0 gives a quadratic in cos(x) . → Factor, or use the quadratic formula to solve for cos(x), and then solve for x.

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