find ( gleft(\frac{pi}{4}\right) ) for ( g(\theta)=cos (3 \theta+pi) ).

Explanation:

Step1: Differentiate (g(\theta)) using chain rule

Let (u = 3\theta+\pi), then (g(\theta)=\cos(u)). The derivative of (\cos(u)) with respect to (u) is (-\sin(u)), and the derivative of (u = 3\theta+\pi) with respect to (\theta) is (3). By the chain rule (\frac{dg}{d\theta}=\frac{dg}{du}\cdot\frac{du}{d\theta}), so (g'(\theta)=- 3\sin(3\theta+\pi)).

Step2: Substitute (\theta=\frac{\pi}{4}) into (g'(\theta))

(g'(\frac{\pi}{4})=-3\sin(3\times\frac{\pi}{4}+\pi)). Simplify the argument of the sine function: (3\times\frac{\pi}{4}+\pi=\frac{3\pi}{4}+\pi=\frac{3\pi + 4\pi}{4}=\frac{7\pi}{4}). Since (\sin(\frac{7\pi}{4})=-\frac{\sqrt{2}}{2}), then (g'(\frac{\pi}{4})=-3\times(-\frac{\sqrt{2}}{2})).

Answer:

(\frac{3\sqrt{2}}{2})