# derivative of cos^3(3x)?

• Oct 6th 2012, 12:06 PM
daniels
derivative of cos^3(3x)?
is it -3sin^3(3x) or did i miss something?
• Oct 6th 2012, 12:11 PM
TheEmptySet
Re: derivative of cos^3(3x)?
Quote:

Originally Posted by daniels
is it -3sin^3(3x) or did i miss something?

Yes you need to use the chain rule. Twice.

let

$f(u)=u^3$

$u=cos(v)$

and

$v=3x$

Then by the chain rule you get

$\frac{df}{dx}=\frac{df}{du}\frac{du}{dv}\frac{dv}{ dx}$

$3u^2(-\sin(v))(3)=-9\cos^2(3x)\sin(x)$
• Oct 6th 2012, 12:16 PM
daniels
Re: derivative of cos^3(3x)?
so then im guessing for f(x) = 2sin(2x) + e^(3x^2) + cos^3(3x)

f '(x) isn't= 4cos(2x) + 3e^(x^2) + -3sin^3(3x) ????
• Oct 6th 2012, 12:19 PM
TheEmptySet
Re: derivative of cos^3(3x)?
Quote:

Originally Posted by daniels
so then im guessing for f(x) = 2sin(2x) + e^(3x^2) + cos^3(3x)

f '(x) isn't= 4cos(2x) + 3e^(x^2) + -3sin^3(3x) ????

The derivative of the first term is correct, you need to use the chain rule on the last two terms.
• Oct 6th 2012, 12:29 PM
daniels
Re: derivative of cos^3(3x)?
okay thnks