Find y' for y = sqrt((1-x)/(1+x))

Find $y'$ for $y = \sqrt{\dfrac{1-x}{1+x}}$.

Step 1: Apply the chain rule

Write $y = u^{1/2}$ with $u = \dfrac{1-x}{1+x}$, so that

\[ y' = \frac{1}{2\sqrt{u}}\cdot u'. \]

Step 2: Differentiate the inner quotient

With numerator $1-x$ (derivative $-1$) and denominator $1+x$ (derivative $1$): \[u' = \frac{(-1)(1+x) - (1-x)(1)}{(1+x)^{2}} = \frac{-1-x-1+x}{(1+x)^{2}} = \frac{-2}{(1+x)^{2}}.\]

Step 3: Substitute

Note that $\dfrac{1}{\sqrt{u}} = \sqrt{\dfrac{1+x}{1-x}}$, the reciprocal radical: \[y' = \frac{1}{2}\sqrt{\frac{1+x}{1-x}}\cdot\frac{-2}{(1+x)^{2}} = -\frac{1}{(1+x)^{2}}\cdot\frac{\sqrt{1+x}}{\sqrt{1-x}}.\]

Step 4: Simplify the powers of $1+x$

Since $\dfrac{\sqrt{1+x}}{(1+x)^{2}} = (1+x)^{\frac12 - 2} = (1+x)^{-3/2}$: $y' = -\frac{1}{(1+x)^{3/2}\sqrt{1-x}}$.

Step 5: Verification

\[x = 0:\quad y = 1,\quad y' = -\frac{1}{1\cdot 1} = -1; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx -1.0 \;\;\checkmark\]

\[x = \tfrac12:\quad y' = -\frac{1}{(1.5)^{3/2}\sqrt{0.5}} \approx -\frac{1}{1.83712\cdot 0.70711} \approx -0.76980 \;\;\checkmark\]