<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] Alternative approach:
Start out differently: Let
$$A = \frac{1-x}{1+x}, ~B = A^{(1/2)}.$$
Then
$$\frac{d [B]}{dx} = \frac{1}{2} \times {A}^{-(1/2)} \times \frac{dA}{dx}. \tag1 $$
and
$$\frac{1}{2} \times {A}^{-(1/2)} = \frac{1}{2} \times \sqrt{\frac{1+x}{1-x}}.\tag2 $$
So, what remains is to compute $~\displaystyle \frac{dA}{dx},~$ combine that with (2) above, and then try to simplify it.
$$\frac{dA}{dx} = \frac{d\left[\frac{1-x}{1+x}\right]}{dx} $$
$$= \frac{[(1 + x)(-1)] - [(1-x)(1)]}{(1 + x)^2} = \frac{-2}{(1 + x)^2} \tag3$$
Combining (2) and (3) results in
$$(-1) \times (1+x)^{(1/2)} \times (1-x)^{(-1/2)} \times (1+x)^{(-2)}$$
$$ = (-1) \times (1-x)^{(-1/2)} \times (1 + x)^{(-3/2)}$$
$$= (-1) (1+x)^{(-1)} \times [(1+x)(1 - x)]^{(-1/2)}$$
$$= \frac{-1}{1+x} \times \frac{1}{\sqrt{1 - x^2}}.$$
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