Differentiate with respect to $x$:

1. Express the exponent in index form to prepare for differentiation.

Let $y = e^{\frac{4}{\sqrt{x}}}$.
Rewrite the exponent: $\frac{4}{\sqrt{x}} = 4x^{-\frac{1}{2}}$
So, $y = e^{4x^{-\frac{1}{2}}}$.

2. Apply the chain rule to differentiate the exponential function.

$\frac{dy}{dx} = e^{4x^{-\frac{1}{2}}} \cdot \frac{d}{dx}\left(4x^{-\frac{1}{2}}\right)$

3. Differentiate the exponent using the power rule and simplify.

$\frac{d}{dx}\left(4x^{-\frac{1}{2}}\right) = 4 \left(-\frac{1}{2}\right)x^{-\frac{3}{2}} = -2x^{-\frac{3}{2}}$

Therefore,
$\frac{dy}{dx} = -2x^{-\frac{3}{2}} e^{4x^{-\frac{1}{2}}} = -\frac{2}{x^{\frac{3}{2}}} e^{\frac{4}{\sqrt{x}}}$

1. Simplify the expression using the laws of logarithms to make differentiation easier

Let $y = \ln\left(\frac{\sin x}{2 - \sin x}\right)^2$.
Using the power and quotient rules of logarithms:
$y = 2 \ln\left(\frac{\sin x}{2 - \sin x}\right)$
$y = 2 [\ln(\sin x) - \ln(2 - \sin x)]$

2. Differentiate the simplified expression with respect to $x$ using the chain rule

$\frac{dy}{dx} = 2 \left[ \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) - \frac{1}{2 - \sin x} \cdot \frac{d}{dx}(2 - \sin x) \right]$
$\frac{dy}{dx} = 2 \left[ \frac{\cos x}{\sin x} - \frac{-\cos x}{2 - \sin x} \right]$
$\frac{dy}{dx} = 2 \left[ \frac{\cos x}{\sin x} + \frac{\cos x}{2 - \sin x} \right]$

3. Combine the terms into a single simplified fraction

$\frac{dy}{dx} = 2 \cos x \left[ \frac{1}{\sin x} + \frac{1}{2 - \sin x} \right]$
$\frac{dy}{dx} = 2 \cos x \left[ \frac{(2 - \sin x) + \sin x}{\sin x(2 - \sin x)} \right]$
$\frac{dy}{dx} = 2 \cos x \left[ \frac{2}{\sin x(2 - \sin x)} \right]$
$\frac{dy}{dx} = \frac{4 \cos x}{\sin x(2 - \sin x)}$