To find the gradient of the curve, we need to differentiate \(y = \frac{\sin x}{1 + \cos x}\) with respect to \(x\).

Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = \sin x\) and \(v = 1 + \cos x\).

Then, \(\frac{du}{dx} = \cos x\) and \(\frac{dv}{dx} = -\sin x\).

Substituting these into the quotient rule gives:

\(\frac{dy}{dx} = \frac{(1 + \cos x)(\cos x) - (\sin x)(-\sin x)}{(1 + \cos x)^2}\)

Simplifying the numerator:

\((1 + \cos x)\cos x + \sin^2 x = \cos x + \cos^2 x + \sin^2 x\)

Using the identity \(\cos^2 x + \sin^2 x = 1\), the numerator becomes:

\(\cos x + 1\)

Thus, \(\frac{dy}{dx} = \frac{1}{1 + \cos x}\).

Since \(-1 < \cos x < 1\) for \(-\pi < x < \pi\), \(1 + \cos x > 0\).

Therefore, \(\frac{1}{1 + \cos x} > 0\), showing that the gradient is positive for all \(x\) in the interval.