Question: I've been stuck trying to differentiate this expression for a while now: 

$$\sqrt{\frac{1-2x}{1+2x}}$$

This is what I've tried so far:

first I split the square root

$$\frac{\sqrt{1-2x}}{\sqrt{1+2x}}$$

then tried to apply the quotient rule 

$$\frac{d\frac{f}{g}}{dx}=\frac{\frac{df}{dx}g-f\frac{dg}{dx}}{g^2}$$

and got:

$$\dfrac{\frac{-1}{\sqrt{1-2x}}\cdot\sqrt{1+2x}-\sqrt{1-2x}\cdot\frac{1}{\sqrt{1+2x}}}{(\sqrt{1+2x})^2}$$

This is as far as I've gotten and I'm not sure what to do next.

Answer: 
There are several ways to perform the differentiation. Let $$f(x) = \sqrt{\frac{1-2x}{1+2x}}.$$ Then:

**Method 1. Polynomial long division with chain rule.**

Write $$\frac{1 - 2x}{1 + 2x} = -\frac{2x + 1 - 2}{2x + 1} = -1 + \frac{2}{2x+1}.$$ Then by the chain rule, $$\frac{df}{dx} = \frac{1}{2}\left(\frac{1-2x}{1+2x}\right)^{-1/2} \frac{d}{dx}\left[-1 + \frac{2}{2x+1}\right] = \frac{1}{2}\left(\frac{1 + 2x}{1 - 2x}\right)^{1/2} \cdot 2(-1)(2x+1)^{-2} \cdot 2.$$ Now simplifying, we get $$\frac{df}{dx} = \frac{-2}{(1-2x)^{1/2}(1 + 2x)^{3/2}}.$$

**Method 2. Logarithmic differentiation.**
Write $$\log f(x) = \frac{1}{2} \left( \log(1 - 2x) - \log(1 + 2x)\right).$$ Then $$\frac{d}{dx} \left[\log f(x)\right] = \frac{1}{f(x)} \frac{df}{dx} = \frac{1}{2} \left( \frac{-2}{1 - 2x} - \cdot \frac{2}{1 + 2x} \right) = -\frac{2}{(1-2x)(1+2x)}.$$ Therefore $$\frac{df}{dx} = -\frac{2f(x)}{(1-2x)(1+2x)} = -\frac{2}{(1-2x)^{1/2}(1+2x)^{3/2}}.$$

**Method 3. Quotient rule.**
With the choice $$p(x) = \sqrt{1-2x}, \quad q(x) = \sqrt{1+2x},$$ from which we obtain $$p'(x) = -(1-2x)^{-1/2}, \quad q(x) = (1+2x)^{-1/2},$$ we compute
$$\frac{df}{dx} = \frac{d}{dx}\left[\frac{p(x)}{q(x)}\right] = \frac{q(x)p'(x) - p(x)q'(x)}{q(x)^2} = \frac{-(1+2x)^{1/2}(1-2x)^{-1/2} - (1-2x)^{1/2}(1+2x)^{-1/2}}{(1+2x)} = -\frac{(1+2x) + (1-2x)}{(1+2x)(1-2x)^{1/2}(1+2x)^{1/2}} = -\frac{2}{(1-2x)^{1/2}(1+2x)^{3/2}},$$ where in the penultimate step we had multiplied the numerator and denominator by $(1-2x)^{1/2}(1+2x)^{1/2}$.

There are other methods as described in other answers, such as employing substitutions or implicit differentiation, but they are covered quite well and so these need not be repeated here.
For this type, a trigonometric substitution, $\color{green}{2x= \cos(2\theta)}$ will be helpful. ($0<\theta<\pi/2$)

So,

$$y = \sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}} = \sqrt{\tan^2\theta} = \tan\theta$$

$$\frac{dy}{dx} = \frac{d}{d\theta}\tan\theta\cdot\frac{d\theta}{dx} = \sec^2\theta\cdot\frac{-1}{\sin(2\theta)} = -\frac{1}{\cos^2\theta\sin(2\theta)} = -\frac{1}{\frac{1+\cos(2\theta)}{2}\cdot\sqrt{1-\cos^2(2\theta)}}$$

[sine is positive in first and second quadrants ($0<2\theta<\pi$)]

$$\color{blue}{\frac{dy}{dx} = -\frac{2}{(1+2x)\sqrt{1-4x^2}}}$$