​$$\text { If } f(x)=\frac{1-x}{2+x} \text {, then find } f^{\prime}(x) \text {. }$$​​

Correct option is B

​$$\begin{aligned}&\text{The Quotient Rule states that if you have a function } \frac{u}{v}, \text{ its derivative is:} \\&\quad \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \\[10pt]&\textbf{Step-by-Step Solution:} \\[5pt]&1. \text{Identify } u \text{ and } v: \\&\quad u = 1 - x \\&\quad v = 2 + x \\[5pt]&2. \text{Compute } \frac{du}{dx} \text{ and } \frac{dv}{dx}: \\&\quad \frac{du}{dx} = \frac{d}{dx}(1 - x) = -1 \\&\quad \frac{dv}{dx} = \frac{d}{dx}(2 + x) = 1 \\[5pt]&3. \text{Apply the Quotient Rule:} \\&\quad f'(x) = \frac{(2 + x)(-1) - (1 - x)(1)}{(2 + x)^2} \\[5pt]&4. \text{Simplify the numerator:} \\&\quad f'(x) = \frac{-2 - x - 1 + x}{(2 + x)^2} = \frac{-3}{(2 + x)^2}\end{aligned}$$​​