find the derivative of the function, y = arctan√(1 - x)/(1 + x) y =

Explanation:

Step1: Let $u = \sqrt{\frac{1 - x}{1 + x}}$

$y=\arctan(u)$

Step2: Find the derivative of $y$ with respect to $u$

The derivative of $y = \arctan(u)$ is $y_{u}'=\frac{1}{1 + u^{2}}$

Step3: Find the derivative of $u$ with respect to $x$

First, rewrite $u=\left(\frac{1 - x}{1 + x}\right)^{\frac{1}{2}}$. Using the chain - rule and the quotient - rule. The quotient - rule states that if $v=\frac{1 - x}{1 + x}$, then $v'=\frac{(1 - x)'(1 + x)-(1 - x)(1 + x)'}{(1 + x)^{2}}=\frac{-1(1 + x)-(1 - x)\times1}{(1 + x)^{2}}=\frac{-1 - x - 1 + x}{(1 + x)^{2}}=\frac{-2}{(1 + x)^{2}}$ By the chain - rule, $u'=\frac{1}{2}\left(\frac{1 - x}{1 + x}\right)^{-\frac{1}{2}}\times v'=\frac{1}{2}\sqrt{\frac{1 + x}{1 - x}}\times\frac{-2}{(1 + x)^{2}}=\frac{-1}{\sqrt{(1 - x)(1 + x)^{3}}}$

Step4: Use the chain - rule $y_{x}'=y_{u}'\cdot u'$

Substitute $u = \sqrt{\frac{1 - x}{1 + x}}$ into $y_{u}'$ and multiply by $u'$: $y_{x}'=\frac{1}{1+\frac{1 - x}{1 + x}}\cdot\frac{-1}{\sqrt{(1 - x)(1 + x)^{3}}}$ Simplify $\frac{1}{1+\frac{1 - x}{1 + x}}=\frac{1 + x}{(1 + x)+(1 - x)}=\frac{1 + x}{2}$ So $y_{x}'=\frac{1 + x}{2}\cdot\frac{-1}{\sqrt{(1 - x)(1 + x)^{3}}}=-\frac{1}{2\sqrt{1 - x^{2}}(1 + x)}$

Answer:

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