<s>[INST] 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?
 [/INST] It's easier to use the chain rule, that is, if $h=g\circ f$, then $$\frac{dh}{dx} = \frac{dg}{df}\cdot \frac{df}{dx}$$ or, written in another way, $$(g\circ f)' = g'\circ f\cdot f'$$
In this case, you can take $g$ as the square root funcion that is $g(x)=\sqrt{x}$, and $f$ as $f(x)=\frac{1-x}{1+x}$.
This quickly leads to the solution, since:

*

*The derivative of $g$ is $g'(x)=\frac{1}{2\sqrt{x}}$

*The derivative of $f$ is $$\frac{-1\cdot(1+x)-1\cdot(1-x)}{(1+x)^2} = \frac{-1-x-1+x}{(1+x)^2} = -\frac{2}{(1+x)^2}$$that the derivative is

which means the derivative of the compositum is
$$\frac{1}{2\sqrt{f(x)}}\cdot f'(x) = \frac{\sqrt{1+x}}{2\sqrt{1-x}}\cdot\frac{-2}{(1+x)^2} = -\frac{1}{\sqrt{1-x}\cdot\sqrt{1+x}\cdot(1+x)} =-\frac{1}{\sqrt{1-x^2}\cdot(1+x)}$$

That said, you can also continue from what you wrote. Finding the least common denominator will get you the following:
$$\begin{align}
\frac{dy}{dx}&=-\frac{(1+x)^\frac12}{2(1+x)\sqrt{1-x}}-\frac{(1-x)^\frac12}{2(1+x)\sqrt{1+x}}\\
&=-\frac{\sqrt{1+x}\cdot[(1+x)^\frac12]}{2(1+x)\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}\cdot [(1-x)^\frac12]}{2(1+x)\sqrt{1+x}\sqrt{1-x}}\\
&=-\frac{1-x}{2(1+x)\sqrt{1+x}\sqrt{1-x}} - \frac{1+x}{2(1+x)\sqrt{1+x}\sqrt{1-x}}\\
&=-\frac{1-x+1+x}{2(1+x)\sqrt{(1+x)(1-x)}}\\
&=-\frac{2}{2(1+x)\sqrt{1-x^2}}
\end{align}$$
which is the same result (as it should be :).
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