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: 
$\newcommand{\d}{\mathrm d} $
Let $y=\sqrt{\dfrac{1-2x}{1+2x}}$. So $y^2=\dfrac{1-2x}{1+2x}$. Differentiating, 
$$2y\dfrac{\d y}{\d x}=\dfrac{-2}{1+2x}-\dfrac{2(1-2x)}{(1+2x)^2}\quad (\text {product rule}).$$
Hence $$\begin{aligned}
\dfrac{\d y}{\d x}&=\dfrac{-1}y\left(\dfrac{1}{1+2x}+\dfrac{1-2x}{(1+2x)^2}\right)\\&=\dfrac{-1}{y(1+2x)}\left(\dfrac{1+2x+1-2x}{1+2x}\right)\\&=\dfrac{-2}{y(1+2x)^2}\\&=\sqrt{\dfrac{1+2x}{1-2x}}\cdot\dfrac{-2}{(1+2x)^2}\\&=\dfrac{-2}{(1+2x)\sqrt{1-4x^2}}.\end{aligned}$$
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 ($0<2\theta<\pi$)]

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