Find $\displaystyle f'(x_0)$, when $\displaystyle f'(x)=2sin(2x)+3cosx+\frac{\pi }{2},$ $\displaystyle x_{0}=\frac{\pi}{6}$

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I think...

$\displaystyle f'(x)=(2sin(2x)+3cosx+\frac{\pi }{2})'=(2sin(2x))'+(3cosx)'+(\frac{\pi}{2})'=$

$\displaystyle =2*cos2x-3sinx=...$?