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?
 A: I will directly follow your step.
$$
\begin{aligned}
\frac{\operatorname{d}y}{\operatorname{d}x}&=-\frac{1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{(1+x)^3}}\\
&=-\frac{1}{2}\frac{1}{\sqrt{1+x}}\Big(\frac{1}{\sqrt{1-x}}+\frac{\sqrt{1-x}}{1+x}\Big)\\
&=-\frac{1}{2\sqrt{1+x}}\frac{2}{(1+x)\sqrt{1-x}}\\
&=-\frac{1}{(1+x)\sqrt{1-x^2}}
\end{aligned}
$$
A: Alternative:
Substitute $x=\cos 2u$. This means that $\dfrac{dx}{du} =-2\sin 2u$ and $\dfrac{du}{dx}=-\dfrac{1}{2\sin 2u}$. Then $$y=\sqrt{\frac{1-\cos 2u}{1+\cos 2u}}=\tan u$$
and $$\frac{dy}{dx}=\sec^2u \dfrac{du}{dx}$$$$=-\frac{1}{2\cos^2 u\sin 2u}$$$$=-\frac{1}{(1+\cos 2u)\sqrt{1-\cos^2 2u}}$$$$=-\frac{1}{(1+x)\sqrt{1-x^2}}$$.
A: Continuing from your second-last step:
$$\frac{dy}{dx}=-\frac{(1+x)^\frac12}{2(1+x)\sqrt{1-x}}-\frac{(1-x)^\frac12}{2(1+x)\sqrt{1+x}}$$
$$=-\frac{\sqrt{1+x}^2}{2(1+x)\sqrt{1-x}\sqrt{1+x}}-\frac{\sqrt{1-x}^2}{2(1+x)\sqrt{1+x}\sqrt{1-x}}$$
$$=\frac{-(1+x)-(1-x)}{2(1+x)\sqrt{1-x^2}}=\frac{-1}{(1+x)\sqrt{1-x^2}}$$
A: 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 :).
A: 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
$$(-1) \times (1+x)^{(1/2)} \times (1-x)^{(-1/2)} \times (1+x)^{(-2)}$$
$$ = (-1) \times (1-x)^{(-1/2)} \times (1 + x)^{(-3/2)}$$
$$= (-1) (1+x)^{(-1)} \times [(1+x)(1 - x)]^{(-1/2)}$$
$$= \frac{-1}{1+x} \times \frac{1}{\sqrt{1 - x^2}}.$$
