Originally Posted by
Archie Meade hi flipboi,
The derivative of $\displaystyle Cosx$ is $\displaystyle -Sinx$,
so the derivative of $\displaystyle 2Cosx$ is the derivative of $\displaystyle (Cosx+Cosx)$ which is $\displaystyle (-Sinx-Sinx)=-2Sinx$.
If you use the trigonometric identity $\displaystyle Cos^2x=\frac{1}{2}(1+Cos2x)$ you can more easily find the derivative of $\displaystyle Cos^2x.$
$\displaystyle \frac{d}{dx}\frac{1}{2}(1+Cos2x)=\frac{1}{2}\frac{ d}{dx}Cos2x$
since the derivative of a constant is zero.
Now you have to use the Chain Rule to complete.
$\displaystyle u=2x,\ \frac{du}{dx}=2$
$\displaystyle \frac{d}{dx}Cos2x=\frac{d}{dx}Cosu=\frac{du}{dx}\f rac{d}{du}Cosu=2(-Sinu)=-2Sin2x$
Therefore the derivative of $\displaystyle Cos^2x$ is $\displaystyle -Sin2x$
Combine the two for your final answer.