Derivative of a linear combination of trig functions, evaluated at a point

Differentiate each term separately using linearity: $\frac{d}{dx}[-2\cos x] = -2(-\sin x) = 2\sin x$, $\frac{d}{dx}[9\sin x] = 9\cos x$, and $\frac{d}{dx}[-5] = 0$. Adding these gives $g'(x) = 2\sin x + 9\cos x$.

Now substitute $x = \pi/2$: $\sin(\pi/2) = 1$ and $\cos(\pi/2) = 0$, so $g'(\pi/2) = 2(1) + 9(0) = 2$.