\(\frac{x}{x + 1}\)
\(\frac{-1}{x + 1}\)
\(\frac{1 - x}{(x + 1)^2}\)
\(\frac{1}{(x + 1)^2}\)
Correct answer is D
\(\frac{d}{dx}(\frac{x }{x + 1}\)) = \(\frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\)
u = x, \(\frac{du}{dx}\) = 1, v = x + 1, \(\frac{dv}{dx}\) = 1
= \(\frac{(x + 1)(1) - x (1)}{(x + 1)^2}\)
= \(\frac{x + 1 - x}{(x + 1)^2}\)
= \(\frac{1}{(x + 1)^2}\)