The given function is \( \displaystyle y = \sqrt{\frac{1\, – x}{1 + x}} \) and we need to find the derivative of this function with respect to \( x \).
Solution:
The given function is:
\( \begin{aligned} y &= \sqrt{\frac{1 – x}{1 + x}} \\ &= \left( \frac{1 – x}{1 + x} \right)^{\frac{1}{2}} \\ \end{aligned} \)Let \(\displaystyle u = \frac{1 – x}{1 + x} \), so: \(\displaystyle y = u^{\frac{1}{2}} \)
Differentiate \( y \) with respect to \( u \):
\( \begin{aligned} \frac{dy}{du} = \frac{1}{2} u^{-\frac{1}{2}} = \frac{1}{2\sqrt{u}} \\ \end{aligned} \)Now, differentiate \( u \) with respect to \( x \) using the quotient rule:
\( \begin{aligned} \frac{du}{dx} &= \frac{(1 + x)(-1) – (1 – x)(1)}{(1 + x)^2} \\ &= \frac{-1 – x – 1 + x}{(1 + x)^2} = \frac{-2}{(1 + x)^2} \\ \end{aligned} \)Multiply \(\frac{dy}{du} \) and \(\frac{du}{dx}\) to find \(\frac{dy}{dx}\):
\( \begin{aligned} \frac{dy}{dx} &= \frac{dy}{du} \cdot \frac{du}{dx} \\ &= \frac{1}{2\sqrt{u}} \cdot \left( \frac{-2}{(1 + x)^2} \right) \\ &= \frac{-1}{\sqrt{u} (1 + x)^2} \\ &= \frac{-1}{\sqrt{\frac{1 – x}{1 + x}} (1 + x)^2} \\ &= \frac{-\sqrt{1 + x}}{\sqrt{1 – x} (1 + x)^2} \\ &= -\frac{1}{(1 + x) \sqrt{(1 – x)(1+x)}} \\ &= -\frac{1}{(1 + x) \sqrt{1 – x^2}} \\ \end{aligned} \)Thus, the derivative of the function is:
\( \displaystyle \boxed{\frac{dy}{dx} = -\frac{1}{(1 + x) \sqrt{1 – x^2}} } \)Please let me know in the comments if you find any errors in this solution.