let ( f(x)=1 + 3cos2x ), for ( 0leq xleqpi ).\n(a) find ( f(x) ). 2 marks

Explanation:

Step1: Differentiate each term

Differentiate (1) and (3\cos(2x)) separately. The derivative of a constant (c) is (0), so (\frac{d}{dx}(1) = 0). For (y = 3\cos(2x)), use the chain - rule (\frac{d}{dx}(u(v(x)))=u^{\prime}(v(x))\cdot v^{\prime}(x)). Let (u(v)=\cos(v)) and (v = 2x). Then (u^{\prime}(v)=-\sin(v)) and (v^{\prime}(x) = 2). So (\frac{d}{dx}(3\cos(2x))=3\times(-\sin(2x))\times2).

Step2: Combine the derivatives

(f^{\prime}(x)=\frac{d}{dx}(1)+\frac{d}{dx}(3\cos(2x))). Substitute the derivatives from Step 1: (f^{\prime}(x)=0 - 6\sin(2x)).

Answer:

(f^{\prime}(x)=-6\sin(2x))