what it the derivative of $\displaystyle (cosX)^3 $ using rules
Is it as easy as $\displaystyle 3(-sinX)2 =-3sin^2X$ ?
You need to use the chain rule here
Find $\displaystyle (\cos^3x)'$
Let's make $\displaystyle y=\cos^3x$ and $\displaystyle u=\cos x\implies y=u^3$
So $\displaystyle \frac{dy}{du} = 3u^2$ and $\displaystyle \frac{du}{dx} = -\sin x$
By the chain rule $\displaystyle \frac{dy}{dx}= \frac{dy}{du}\times \frac{du}{dx} $
Can you finish it?