# Homework Help: Derivative with Constant

1. Apr 7, 2010

### Hockeystar

1. The problem statement, all variables and given/known data

Find the max/min of the function for [0,pi/2]

f(x)= kcos2xsin(x)

2. Relevant equations

Product Rule and chain rule

3. The attempt at a solution

f'(x)= k(2cosx(-sinx)(sinx)+cos2xcosx)

f'(x)= kcosx(-2sin2x+cos2x)

Set eqaution to 0

(2sin2x+cos(x)2) = 0

2(1-cos2x)+cos(x)2 = 0

cos2x = 2
?
Now I'm stuck. There should be a max or min.

2. Apr 7, 2010

### ystael

You appear to have made two errors here. You neglected to account for the other factor, that perhaps $$\cos x = 0$$. You also seem to have dropped a minus sign; if the second factor is zero, you should have $$-2 \sin^2 x + \cos^2 x = 0$$.

You may have an easier time with this second factor if you rewrite it so that only one trig function of $$x$$ appears.

3. Apr 7, 2010

### Hockeystar

Man I hate when I forget a negative sign. I also scrwed up the interval is supposed to be (0,pi/2) instead of [0,pi/2]. Thank you for pointing that out.