16. $\\frac { d } { d x } \\left( \\frac { x + 1 } { x ^ { 2 } + 1 } \\right) =$ a $\\frac { x ^ { 2 } + 2 x…

16. $\\frac { d } { d x } \\left( \\frac { x + 1 } { x ^ { 2 } + 1 } \\right) =$ a $\\frac { x ^ { 2 } + 2 x - 1 } { ( x ^ { 2 } + 1 ) ^ { 2 } }$ b $\\frac { - x ^ { 2 } - 2 x + 1 } { x ^ { 2 } + 1 }$ c $\\frac { - x ^ { 2 } - 2 x + 1 } { ( x ^ { 2 } + 1 ) ^ { 2 } }$ d $\\frac { 3 x ^ { 2 } + 2 x + 1 } { ( x ^ { 2 } + 1 ) ^ { 2 } }$ e $\\frac { 1 } { 2 x }$

16. $\\frac { d } { d x } \\left( \\frac { x + 1 } { x ^ { 2 } + 1 } \\right) =$ a $\\frac { x ^ { 2 } + 2 x - 1 } { ( x ^ { 2 } + 1 ) ^ { 2 } }$ b $\\frac { - x ^ { 2 } - 2 x + 1 } { x ^ { 2 } + 1 }$ c $\\frac { - x ^ { 2 } - 2 x + 1 } { ( x ^ { 2 } + 1 ) ^ { 2 } }$ d $\\frac { 3 x ^ { 2 } + 2 x + 1 } { ( x ^ { 2 } + 1 ) ^ { 2 } }$ e $\\frac { 1 } { 2 x }$

Answer

Explanation:

Step1: Apply the quotient rule

The quotient rule states that if (y = \frac{u}{v}), then (y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}). Here, (u=x + 1) and (v=x^{2}+1). First, find (u^\prime) and (v^\prime). (u^\prime=\frac{d}{dx}(x + 1)=1) (v^\prime=\frac{d}{dx}(x^{2}+1)=2x)

Step2: Substitute into the quotient rule formula

[ \begin{align*} \frac{d}{dx}\left(\frac{x + 1}{x^{2}+1}\right)&=\frac{(1)\times(x^{2}+1)-(x + 1)\times(2x)}{(x^{2}+1)^{2}}\ &=\frac{x^{2}+1-(2x^{2}+2x)}{(x^{2}+1)^{2}}\ &=\frac{x^{2}+1 - 2x^{2}-2x}{(x^{2}+1)^{2}}\ &=\frac{-x^{2}-2x + 1}{(x^{2}+1)^{2}} \end{align*} ]

Answer:

C. (\frac{-x^{2}-2x + 1}{(x^{2}+1)^{2}})