(32) If \( y=\cos (\pi-3 x)^{2} \), then \( \frac{d y}{d x}= \) \( \begin{array}{ll}\text { (a) }-6 \cos (\pi-3 x) & \text { (b) }-\sin (\pi-3 x)^{2}+2(\pi-3 x) \cos (\pi-3 x)^{2} \\ \text { (c) } 6(\pi-3 x) \sin (\pi-3 x)^{2} & \text { (d) }-2 \sin (\pi-3 x)^{2}\end{array} \)

To find \(\frac{dy}{dx}\) when \( y = \cos((\pi - 3x)^2) \), we'll apply the chain rule. First, we need to take the derivative of \(\cos(u)\) where \(u = (\pi - 3x)^2\): \[ \frac{dy}{du} = -\sin(u). \] Next, we need to differentiate \(u = (\pi - 3x)^2\): \[ \frac{du}{dx} = 2(\pi - 3x)(-3) = -6(\pi - 3x). \] Now, applying the chain rule: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\sin((\pi - 3x)^2) \cdot (-6(\pi - 3x)) = 6(\pi - 3x) \sin((\pi - 3x)^2). \] Hence, the correct answer is: (c) \( 6(\pi - 3x) \sin((\pi - 3x)^2) \).