6. 0 / 1 points\nfind the derivative of the function.\n$y = \\arctan(\\sqrt{\\frac{1 - x}{1 + x}})$\n$y =…

Explanation:

Step1: Apply the chain rule

The chain rule states that if (y = f(g(x))), then (y'=f'(g(x))\cdot g'(x)). Here, (f(u)=\arctan(u)) and (u = g(x)=\sqrt{\frac{1 - x}{1 + x}}). First, find (f'(u)). Since (\frac{d}{du}(\arctan(u))=\frac{1}{1 + u^{2}}), then (f'(g(x))=\frac{1}{1+(\sqrt{\frac{1 - x}{1 + x}})^{2}}).

Simplify (1+(\sqrt{\frac{1 - x}{1 + x}})^{2}): [ \begin{align*} 1+\frac{1 - x}{1 + x}&=\frac{(1 + x)+(1 - x)}{1 + x}\ &=\frac{1+x + 1 - x}{1 + x}\ &=\frac{2}{1 + x} \end{align*} ] So (f'(g(x))=\frac{1 + x}{2}).

Step2: Find (g'(x))

Let (u=\frac{1 - x}{1 + x}), so (g(x)=\sqrt{u}=u^{\frac{1}{2}}). First, use the quotient rule to find (\frac{du}{dx}). The quotient rule (\frac{d}{dx}(\frac{v}{w})=\frac{v'w - vw'}{w^{2}}), where (v = 1 - x), (v'=-1), (w = 1 + x), (w' = 1). Then (\frac{du}{dx}=\frac{-1\cdot(1 + x)-(1 - x)\cdot1}{(1 + x)^{2}}=\frac{-1 - x-1 + x}{(1 + x)^{2}}=\frac{-2}{(1 + x)^{2}}).

By the chain rule, (g'(x)=\frac{dg}{du}\cdot\frac{du}{dx}). Since (\frac{dg}{du}=\frac{1}{2}u^{-\frac{1}{2}}), then (g'(x)=\frac{1}{2}(\frac{1 - x}{1 + x})^{-\frac{1}{2}}\cdot\frac{-2}{(1 + x)^{2}}).

Simplify (g'(x)): [ \begin{align*} g'(x)&=\frac{1}{2}\sqrt{\frac{1 + x}{1 - x}}\cdot\frac{-2}{(1 + x)^{2}}\ &=-\frac{1}{\sqrt{(1 - x)(1 + x)}(1 + x)} \end{align*} ]

Step3: Calculate (y')

By the chain rule (y'=f'(g(x))\cdot g'(x)). Substitute (f'(g(x))=\frac{1 + x}{2}) and (g'(x)=-\frac{1}{\sqrt{(1 - x)(1 + x)}(1 + x)}) into the formula: [ \begin{align*} y'&=\frac{1 + x}{2}\cdot\left(-\frac{1}{\sqrt{(1 - x)(1 + x)}(1 + x)}\right)\ &=-\frac{1}{2\sqrt{1 - x^{2}}} \end{align*} ]

Answer:

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