If f(x)=xx+1f(x) = \dfrac{x}{x+1}f(x)=x+1x, then f′(x)=f'(x) =f′(x)=A−1(x+1)2\dfrac{-1}{(x+1)^2}(x+1)2−1Bx(x+1)2\dfrac{x}{(x+1)^2}(x+1)2xC1(x+1)2\dfrac{1}{(x+1)^2}(x+1)21check_circleD111ExplanationQuotient rule: 1⋅(x+1)−x⋅1(x+1)2=1(x+1)2\dfrac{1 \cdot (x+1) - x \cdot 1}{(x+1)^2} = \dfrac{1}{(x+1)^2}(x+1)21⋅(x+1)−x⋅1=(x+1)21.