Trig identities and differentiation: I differentiated (cos(3x))^2 and got -6cos(3x)sin(3x) but answer key says -3cos(6x)

[johnflynn]

I have to differentiate (cos (3x))^2 which I did and got -6cos(3x)sin(3x), but the answer says it's -3cos(6x).....please help! Thanks

 

[blamocur]

but the answer says it's -3cos(6x).

Not -3 sin (6x) ?

 

[MaxWong]

It should be $-3 \sin(6x)$

 

[Dr.Peterson]

It should be $-3 \sin(6x)$

They got your answer, and then applied a double-angle identity to get their form, -3sin(6x). Both are correct answers.

 

[MaxWong]

They got your answer, and then applied a double-angle identity to get their form, -3sin(6x). Both are correct answers.

Oh, I meant the answer should not be $-3\cos(6x)$.

 

[Steven G]

If you think that your answer is correct, then use some type of software to verify that [-6cos(3x)sin(3x)] - [-3cos(6x)] = 0.
The fact that the angle doubled, by going from 3x to 6x, you should think double angle formula.

 

[Dr.Peterson]

Oh, I meant the answer should not be $-3\cos(6x)$.

I was confusing you with the OP. I'd like to know why they said the book gave a wrong answer. My "your" referred to the OP:

I have to differentiate (cos (3x))^2 which I did and got -6cos(3x)sin(3x), but the answer says it's -3cos(6x).....please help! Thanks

 

[The Highlander]

Just to clarify, everyone is now agreed that the book's answer is wrong, yes?

The correct answer should be -3sin(6x) not -3cos(6x)
(because $\displaystyle \tiny{\frac{\mathrm{d}}{\mathrm{d}x}}\footnotesize (cos(3x))^2=-6cos(3x)sin(3x)=-3sin(6x)$)