Differentiate 8ecos2x w.r.t x.

The correct option is -16 sin2x e^cos2x.

To differentiate the given function with respect to $x$, we will use the chain rule of differentiation. Let the function be represented as:
$$y=8e^{\cos \left(2x\right)}$$

First, we apply the constant multiple rule of differentiation, which allows us to pull out the constant coefficient:
$$\frac{dy}{dx}=8\cdot \frac{d}{dx}\left(e^{\cos \left(2x\right)}\right)$$

Next, we apply the chain rule to differentiate the exponential function $e^{u\left(x\right)}$, where $u\left(x\right)=\cos \left(2x\right)$. The derivative of $e^u$ with respect to $x$ is $e^u\cdot \frac{du}{dx}$:
$$\frac{d}{dx}\left(e^{\cos \left(2x\right)}\right)=e^{\cos \left(2x\right)}\cdot \frac{d}{dx}(\cos (2x\left)\right)$$

Now, we find the derivative of the trigonometric part, $\cos \left(2x\right)$, by applying the chain rule once again:
$$\frac{d}{dx}(\cos (2x\left)\right)=-\sin \left(2x\right)\cdot \frac{d}{dx}\left(2x\right)$$
Since the derivative of $2x$ is $2$, we have:
$$\frac{d}{dx}(\cos (2x\left)\right)=-2\sin \left(2x\right)$$

Substituting this derivative back into our primary expression yields:
$$\frac{dy}{dx}=8\cdot (e^{\cos \left(2x\right)}\cdot (-2\sin \left(2x\right)\left)\right)$$
Simplifying the constant terms by multiplying $8$ and $-2$ gives the final derivative:
$$\frac{dy}{dx}=-16\sin \left(2x\right)e^{\cos \left(2x\right)}$$