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

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

Step 1: Combine the radicals

A quotient of square roots is the square root of the quotient: $y = \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$.

This converts a quotient-of-roots problem into a single chain-rule problem, exactly the shape of Problem 94 with $\sqrt{x}$ in place of $x$.

Step 2: Apply the chain rule

With $u = \dfrac{1-\sqrt{x}}{1+\sqrt{x}}$ we have $y = u^{1/2}$ and $y' = \dfrac{u'}{2\sqrt{u}}$.

Step 3: Differentiate the inner quotient

Put $p = 1-\sqrt{x}$, $p' = -\dfrac{1}{2\sqrt{x}}$, $q = 1+\sqrt{x}$, $q' = \dfrac{1}{2\sqrt{x}}$. Factoring $-\dfrac{1}{2\sqrt{x}}$ out of the numerator: \[u' = \frac{-\dfrac{1}{2\sqrt{x}}\bigl(1+\sqrt{x}\bigr) - \bigl(1-\sqrt{x}\bigr)\dfrac{1}{2\sqrt{x}}}{\bigl(1+\sqrt{x}\bigr)^{2}} = \frac{-\dfrac{1}{2\sqrt{x}}\Bigl[\bigl(1+\sqrt{x}\bigr)+\bigl(1-\sqrt{x}\bigr)\Bigr]}{\bigl(1+\sqrt{x}\bigr)^{2}} = \frac{-1}{\sqrt{x}\,\bigl(1+\sqrt{x}\bigr)^{2}}.\]

Step 4: Substitute and simplify

Using $\dfrac{1}{\sqrt{u}} = \sqrt{\dfrac{1+\sqrt{x}}{1-\sqrt{x}}}$ and absorbing $\sqrt{1+\sqrt{x}}$ into the square: \[y' = \frac{1}{2}\sqrt{\frac{1+\sqrt{x}}{1-\sqrt{x}}}\cdot\frac{-1}{\sqrt{x}\bigl(1+\sqrt{x}\bigr)^{2}} = -\frac{1}{2\sqrt{x}\,\bigl(1+\sqrt{x}\bigr)^{3/2}\sqrt{1-\sqrt{x}}}.\]

Step 5: Verification

\[x = \tfrac14:\quad \sqrt{x} = \tfrac12 \;\Longrightarrow\; y' = -\frac{1}{2(0.5)(1.5)^{3/2}\sqrt{0.5}} \approx -\frac{1}{1\cdot 1.83712\cdot 0.70711} \approx -0.76980\]

\[\frac{y(0.25+10^{-6})-y(0.25-10^{-6})}{2\times 10^{-6}} \approx -0.76980 \;\;\checkmark\]