The given function is \( \displaystyle y = \sqrt{\frac{x\, – 1}{x + 1}} \) and we need to find the derivative of this function with respect to \( x \).
Solution:
Given function is:
\( \displaystyle y = \sqrt{\frac{x\, – 1}{x + 1}} = \left( \frac{x\, – 1}{x + 1} \right)^{1/2} \)
Let \( \displaystyle u = \frac{x\, – 1}{x + 1} \), so \(\displaystyle y = u^{1/2} \)
Differentiate \(y \) with respect to \( u \),
\(\displaystyle \frac{dy}{du} = \frac{1}{2} u^{-1/2} = \frac{1}{2\sqrt{u}} \)
Differentiate \(u \) with respect to \( x \) using the quotient rule,
\( \displaystyle \frac{du}{dx} = \frac{d}{dx} \left( \frac{x\, – 1}{x + 1} \right) \\ \displaystyle = \frac{(1)(x + 1)\, – (x\, – 1)(1)}{(x + 1)^2} \\ \displaystyle = \frac{2}{(x + 1)^2} \\ \)Now, apply the chain rule:
\( \displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{2}{(x + 1)^2} \\ \displaystyle = \frac{1}{\sqrt{u} (x + 1)^2} = \frac{1}{\sqrt{\frac{x\, – 1}{x + 1}}} \cdot \frac{1}{(x + 1)^2} \\ \displaystyle = \frac{\sqrt{x + 1}}{\sqrt{x\, – 1}} \cdot \frac{1}{(x + 1)^2} \\ \displaystyle = \frac{1}{(x + 1)^{3/2} \sqrt{x\, – 1}} \\ \displaystyle = \frac{1}{(x + 1) \sqrt{x + 1} \sqrt{x\, – 1}} \\ \displaystyle = \frac{1}{(x + 1) \sqrt{(x + 1)(x\, – 1)}} \\ \displaystyle = \frac{1}{(x + 1)\sqrt{x^2\, – 1}} \\ \)Therefore, \( \boxed{\frac{dy}{dx} = \frac{1}{(x + 1)\sqrt{x^2\, – 1}}} \)
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