1. ## derivative of cos^3(3x)?

is it -3sin^3(3x) or did i miss something?

2. ## Re: derivative of cos^3(3x)?

Originally Posted by daniels
is it -3sin^3(3x) or did i miss something?
Yes you need to use the chain rule. Twice.

let

$\displaystyle f(u)=u^3$

$\displaystyle u=cos(v)$

and

$\displaystyle v=3x$

Then by the chain rule you get

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

$\displaystyle 3u^2(-\sin(v))(3)=-9\cos^2(3x)\sin(x)$

3. ## 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) ????

4. ## Re: derivative of cos^3(3x)?

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.

okay thnks

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# derivative of cos^3

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