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 $\frac{dy}{du}$

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

Step3: Find $\frac{du}{dx}$ using the chain - rule and quotient - rule

First, rewrite $u=\left(\frac{1 - x}{1 + x}\right)^{\frac{1}{2}}$. By the chain - rule, $\frac{du}{dx}=\frac{1}{2}\left(\frac{1 - x}{1 + x}\right)^{-\frac{1}{2}}\cdot\frac{d}{dx}\left(\frac{1 - x}{1 + x}\right)$. Using the quotient - rule $\left(\frac{f}{g}\right)'=\frac{f'g - fg'}{g^{2}}$ where $f = 1 - x$, $f'=-1$, $g = 1 + x$, $g' = 1$. $\frac{d}{dx}\left(\frac{1 - x}{1 + x}\right)=\frac{-1\cdot(1 + x)-(1 - x)\cdot1}{(1 + x)^{2}}=\frac{-1 - x - 1 + x}{(1 + x)^{2}}=\frac{-2}{(1 + x)^{2}}$. So $\frac{du}{dx}=\frac{1}{2}\left(\frac{1 + x}{1 - x}\right)^{\frac{1}{2}}\cdot\frac{-2}{(1 + x)^{2}}=\frac{-1}{(1 + x)^{2}}\sqrt{\frac{1 + x}{1 - x}}$

Step4: Use the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

Substitute $u = \sqrt{\frac{1 - x}{1 + x}}$ into $\frac{dy}{du}$ and multiply by $\frac{du}{dx}$: $\frac{dy}{dx}=\frac{1}{1+\frac{1 - x}{1 + x}}\cdot\frac{-1}{(1 + x)^{2}}\sqrt{\frac{1 + x}{1 - x}}$ First, simplify $1+\frac{1 - x}{1 + x}=\frac{1 + x+1 - x}{1 + x}=\frac{2}{1 + x}$. Then $\frac{dy}{dx}=\frac{1 + x}{2}\cdot\frac{-1}{(1 + x)^{2}}\sqrt{\frac{1 + x}{1 - x}}=\frac{-1}{2(1 + x)}\sqrt{\frac{1 + x}{1 - x}}=\frac{-1}{2\sqrt{1 - x^{2}}}$

Answer:

$\frac{-1}{2\sqrt{1 - x^{2}}}$