\begin{exercice*}[\ExerciceRefMethode{TA3_methode_deriv_trigo}]\label{TA3_exo_methode_deriv_trigo}
Soit $f$ une fonction définie sur $\mathbb{R}$. Calculer $f'(x)$.
\begin{colenumerate}{2}
\item $f(x)= x^2+\cos x$
\item $f(x)=\sin 2x$
\item $f(x)=\cos x \sin x$
\item $f(x)= \sin^2 x$
\item $f(x)=x^2\cos x$
\item $f(x)=\cos^2 x$
\item $f(x)=\sin x+\cos x$
\item $f(x)=\dfrac{2\cos x+3}{2\cos x-3}$
\end{colenumerate}
\begin{corrige}
\begin{enumerate}
\item $f'(x)= 2x-\sin x$
\item $f'(x)=2\cos 2x$
\item \begin{tabular}{@{}l}
$f'(x)=-\sin^2x+\cos^2x$ \\
$\phantom{f'(x)}=\cos 2x$
        \end{tabular}
\item \begin{tabular}{@{}l}
$f'(x)= 2\cos x\sin x$ \\
$\phantom{f'(x)}=\sin 2x$
\end{tabular}
\item \begin{tabular}{@{}l}
$f'(x)=2x\cos x-x^2\sin x$\\
$\phantom{f'(x)}=x(2\cos x-x\sin x)$
\end{tabular}
\item \begin{tabular}{@{}l}
$f'(x)=-2\sin x \cos x$\\
$\phantom{f'(x)}=-2\sin 2x$
\end{tabular}
\item $f'(x)=\cos x-\sin x$
\item \begin{tabular}{@{}l}
$f'(x)=\tfrac{12\sin x}{\left(2\cos x-3\right)^2}$
\end{tabular}
\end{enumerate}
\end{corrige}
\end{exercice*}

