Differentiating a trig function

[steve snash]

Homework Statement

Find dy/dx, given
y = sin( 3 x ) cos( 6 x )

Homework Equations

product rule= f'g+g'f

The Attempt at a Solution

I used the product rule and got this, but its supposedly wrong, what have i done wrong?((3*cos(x))*(6*cos(x)))+((6*sin(-x))*(3*sin(x)))

 

[gabbagabbahey]

For starters, $\frac{d}{dx}\sin(3x)=3\cos(3x)\neq3\cos(x)$

 

[Char. Limit]

Here's a general rule that may help you figure it out:

$$\frac{d}{dx}(sin(ax)) = a \times cos(ax)$$

 

[steve snash]

cheers

 

[steve snash]

wtf, it still says its wrong,
((3*cos(3*x))*(6*cos(6*x)))+((6*sin(-6*x))*(3*sin(3*x)))
what am i still doing wrong?

 

[gabbagabbahey]

Where are the extra factors of 6 in your first term and 3 in your second term coming from?

 

[steve snash]

sorry i ment
((3*cos(3*x))*(cos(6*x)))+((6*sin(-6*x))*(sin(3*x)))
but its still wrong, can you simplify this?

 

[gabbagabbahey]

For starters, $\sin(-6x)=-\sin(6x)$...can you think of a trig identity that involves something like $\cos(a)\cos(b)-\sin(a)\sin(b)$?