find $\frac{d^{2}y}{dx^{2}}$ if $y = sqrt{\frac{x - 1}{x + 1}}$

Explanation:

Step1: Rewrite the function

First, rewrite $y = \sqrt{\frac{x - 1}{x+1}}=( \frac{x - 1}{x + 1})^{\frac{1}{2}}$.

Step2: Use the chain - rule and quotient - rule to find the first derivative

The chain - rule states that if $y = u^{\frac{1}{2}}$ and $u=\frac{x - 1}{x + 1}$, then $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and by the quotient - rule $\frac{du}{dx}=\frac{(x + 1)\times1-(x - 1)\times1}{(x + 1)^2}=\frac{x + 1-x + 1}{(x + 1)^2}=\frac{2}{(x + 1)^2}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\frac{1}{2}(\frac{x - 1}{x + 1})^{-\frac{1}{2}}\cdot\frac{2}{(x + 1)^2}=\frac{1}{(x + 1)^2\sqrt{\frac{x - 1}{x + 1}}}=\frac{1}{(x + 1)^{\frac{3}{2}}(x - 1)^{\frac{1}{2}}}$.

Step3: Use the quotient - rule to find the second derivative

The quotient - rule states that if $y=\frac{v}{w}$, then $y^\prime=\frac{v^\prime w - vw^\prime}{w^{2}}$, where $v = 1$, $v^\prime=0$, $w=(x + 1)^{\frac{3}{2}}(x - 1)^{\frac{1}{2}}$. First, find $w^\prime$ using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u=(x + 1)^{\frac{3}{2}}$ and $v=(x - 1)^{\frac{1}{2}}$. $u^\prime=\frac{3}{2}(x + 1)^{\frac{1}{2}}$ and $v^\prime=\frac{1}{2}(x - 1)^{-\frac{1}{2}}$. $w^\prime=\frac{3}{2}(x + 1)^{\frac{1}{2}}(x - 1)^{\frac{1}{2}}+\frac{1}{2}(x + 1)^{\frac{3}{2}}(x - 1)^{-\frac{1}{2}}$. Then $\frac{d^{2}y}{dx^{2}}=\frac{0\times(x + 1)^{\frac{3}{2}}(x - 1)^{\frac{1}{2}}-1\times[\frac{3}{2}(x + 1)^{\frac{1}{2}}(x - 1)^{\frac{1}{2}}+\frac{1}{2}(x + 1)^{\frac{3}{2}}(x - 1)^{-\frac{1}{2}}]}{[(x + 1)^{\frac{3}{2}}(x - 1)^{\frac{1}{2}}]^{2}}$. Simplify the numerator and denominator: [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{-\frac{3}{2}(x + 1)^{\frac{1}{2}}(x - 1)^{\frac{1}{2}}-\frac{1}{2}(x + 1)^{\frac{3}{2}}(x - 1)^{-\frac{1}{2}}}{(x + 1)^{3}(x - 1)}\ &=\frac{-\frac{3(x - 1)+(x + 1)}{2(x - 1)^{\frac{1}{2}}(x + 1)^{\frac{1}{2}}}}{(x + 1)^{3}(x - 1)}\ &=\frac{-(3x-3+x + 1)}{2(x - 1)^{\frac{3}{2}}(x + 1)^{\frac{7}{2}}}\ &=\frac{-(4x - 2)}{2(x - 1)^{\frac{3}{2}}(x + 1)^{\frac{7}{2}}}\ &=\frac{1 - 2x}{(x - 1)^{\frac{3}{2}}(x + 1)^{\frac{7}{2}}} \end{align*} ]

Answer:

$\frac{1 - 2x}{(x - 1)^{\frac{3}{2}}(x + 1)^{\frac{7}{2}}}$