How to differentiate $y=\sqrt{\frac{1-x}{1+x}}$? It is an example question from "Calculus Made Easy" by Silvanus Thompson (page 68-69). He gets to the answer $$\frac{dy}{dx}=-\frac{1}{(1+x)\sqrt{1-x^2}}$$ The differentiation bit of the question is straightforward, but I'm having trouble simplifying it to get the exact answer. My working is the following:
$$y=\sqrt{\frac{1-x}{1+x}}$$
This can be written as $$y=\frac{(1-x)^\frac{1}{2}}{(1+x)^\frac{1}{2}}$$
Using the quotient rule we get $$\frac{dy}{dx}=\frac{(1+x)^\frac12\frac{d(1-x)^\frac12}{dx}-(1-x)^\frac12\frac{d(1+x)^\frac12}{dx}}{1+x}$$
Hence $$\frac{dy}{dx}=-\frac{(1+x)^\frac12}{2(1+x)\sqrt{1-x}}-\frac{(1-x)^\frac12}{2(1+x)\sqrt{1+x}}$$
$$=-\frac{1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{(1+x)^3}}$$
What is the next step?

I will directly follow your step.
$$
\begin{aligned}
\frac{\operatorname{d}y}{\operatorname{d}x}&=-\frac{1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{(1+x)^3}}\\
&=-\frac{1}{2}\frac{1}{\sqrt{1+x}}\Big(\frac{1}{\sqrt{1-x}}+\frac{\sqrt{1-x}}{1+x}\Big)\\
&=-\frac{1}{2\sqrt{1+x}}\frac{2}{(1+x)\sqrt{1-x}}\\
&=-\frac{1}{(1+x)\sqrt{1-x^2}}
\end{aligned}
$$
