From the problem, we have :
[tex]\cos ^2x[/tex]Note that it is the same as :
[tex]\cos ^2x=(\cos x)^2[/tex]And using the general derivative formula :
[tex]\frac{d(u^n)}{dx}=nu^{n-1}du[/tex]The exponent will be multiplied to u raised to n-1 and the derivative of u.
Going back to the problem :
[tex]\frac{d(\cos x)^2}{dx}=2\cos x\cdot d(\cos x)[/tex]Note that the derivative of cos x is -sin x
[tex]\begin{gathered} 2\cos x\cdot d(\cos x)=2\cos x(-\sin x) \\ \Rightarrow-2\cos x\sin x \end{gathered}[/tex]Take note also of the identity :
[tex]\sin 2x=2\sin x\cos x[/tex]So the expression will be :
[tex]-2\sin x\cos x=-\sin 2x[/tex]The answer is -sin 2x
