Derivatives of trigonometric functions

Multiply numerator and denominator by $$\left(\sqrt{1+\sin x} - \sqrt{1-\sin x}\right)$$:

$$f(x) = \frac{\left(\sqrt{1+\sin x} - \sqrt{1-\sin x}\right)^2}{(1+\sin x)-(1-\sin x)}$$

$$\left(\sqrt{1+\sin x} - \sqrt{1-\sin x}\right)^2 = 2 - 2\sqrt{1-\sin^2 x} = 2 - 2\cos x$$

Denominator $$= 2\sin x$$. So:

$$f(x) = \frac{2-2\cos x}{2\sin x} = \frac{1-\cos x}{\sin x}$$

Apply half-angle identities $$1-\cos x = 2\sin^2\dfrac{x}{2}$$ and $$\sin x = 2\sin\dfrac{x}{2}\cos\dfrac{x}{2}$$:

$$f(x) = \frac{2\sin^2\dfrac{x}{2}}{2\sin\dfrac{x}{2}\cos\dfrac{x}{2}} = \tan\frac{x}{2}$$

$$g(x) = \ln\left(\tan\frac{x}{2}\right)$$

$$g'(x) = \frac{1}{\tan\dfrac{x}{2}} \cdot \sec^2\frac{x}{2} \cdot \frac{1}{2} = \frac{1}{2\sin\dfrac{x}{2}\cos\dfrac{x}{2}} = \frac{1}{\sin x} = \cosec x$$

$$g''(x) = \frac{d}{dx}(\cosec x) = \boxed{-\cosec x\cot x}$$