\(\frac{1}{x + 1}\)
\(\frac{1}{(x + 1)^{2}}\)
\(\frac{1 - x}{x + 1}\)
\(\frac{1 - x}{(x + 1)^{2}}\)
Correct answer is B
\(y = \frac{x}{x + 1}\)
Using quotient rule because the function is of the form \(\frac{u(x)}{v(x)}\)
\(\frac{\mathrm d y}{\mathrm d x} = \frac{v\frac{\mathrm d u}{\mathrm d x} - u\frac{\mathrm d v}{\mathrm d x}}{v^{2}}\)
\(\frac{\mathrm d y}{\mathrm d x} = \frac{(x + 1) . 1 - x . 1}{(x + 1)^{2}}\)
= \(\frac{1}{(x + 1)^{2}}\)