The given function is \( \displaystyle y = \frac{1 + \sqrt{x}}{1\, – \sqrt{x}} \) and we need to find the derivative of this function with respect to \( x \).
Solution:
Let \(u = 1 + \sqrt{x} = 1 + x^{1/2} \) and \(v = 1\, – \sqrt{x} = 1\, – x^{1/2} \). Then using the Quotient Rule,
\( \begin{aligned} \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} – u \cdot \frac{dv}{dx}}{v^2} \dots (1) \\ \end{aligned} \)Differentiate \( u \) and \( v \) with respect to \( x \):
\( \begin{aligned} \frac{du}{dx} = \frac{d}{dx}\left(1 + x^{1/2}\right) = 0 + \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \\ \frac{dv}{dx} = \frac{d}{dx}\left(1\, – x^{1/2}\right) = 0\, – \frac{1}{2}x^{-1/2} = -\frac{1}{2\sqrt{x}} \\ \end{aligned} \)Substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into (1).
\( \begin{aligned} \frac{dy}{dx} &= \frac{(1 – \sqrt{x}) \cdot \left(\frac{1}{2\sqrt{x}}\right) – (1 + \sqrt{x}) \cdot \left(-\frac{1}{2\sqrt{x}}\right)}{(1 – \sqrt{x})^2} \\ &= \frac{\frac{1 – \sqrt{x}}{2\sqrt{x}} + \frac{1 + \sqrt{x}}{2\sqrt{x}}}{(1 – \sqrt{x})^2} \\ &= \frac{\frac{(1 – \sqrt{x}) + (1 + \sqrt{x})}{2\sqrt{x}}}{(1 – \sqrt{x})^2} \\ &= \frac{\frac{2}{2\sqrt{x}}}{(1 – \sqrt{x})^2} = \frac{\frac{1}{\sqrt{x}}}{(1 – \sqrt{x})^2} \\ &= \frac{1}{\sqrt{x} (1 – \sqrt{x})^2} \\ \end{aligned} \)Thus, the derivative of the function is:
\( \begin{aligned} \boxed{\frac{dy}{dx} = \frac{1}{\sqrt{x} (1 – \sqrt{x})^2}} \\ \end{aligned} \)Please let me know in the comments if you find any errors in this solution.