We are given $y = 3\sin(-4x)$. This is a composite function: the sine function has inside it $-4x$. We use the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$, where $u = -4x$.

First, differentiate the outer function: if $y = 3\sin u$, then $\frac{dy}{du} = 3\cos u$. The derivative of $\sin u$ is $\cos u$.

Now, differentiate the inner function $u = -4x$: $\frac{du}{dx} = -4$.

Applying the chain rule: $\frac{dy}{dx} = (3\cos u) \cdot (-4) = -12 \cos u$.

Substitute back $u = -4x$: $\frac{dy}{dx} = -12 \cos(-4x)$.

Notice that $\cos(-4x) = \cos(4x)$ because cosine is an even function ($\cos(-\theta) = \cos \theta$). So $-12 \cos(-4x)$ is equivalent to $-12 \cos(4x)$. However, the answer choices list $-12\cos(-4x)$ twice (options A and D are identical), and $12\cos(4x)$ as option B. Since the question asks for $dy/dx$ and we derived $-12 \cos(-4x)$, that is the direct result.

So the correct answer is D) $-12\cos(-4x)$.

Textbook reference: Macmillan Progressive Mathematics 3, pp. 153–154