The given function is
[tex]f(x)=\frac{x+1}{2x^2+2x+3}[/tex]
To find its derivative we should use the rule of division
[tex]\frac{d}{dx}(\frac{u}{v})=\frac{u^{\prime}v-uv^{\prime}}{v^2}[/tex]
Let
[tex]\begin{gathered} u=x+1 \\ v=2x^2+2x+3 \end{gathered}[/tex]
Let us find u' and v' first
[tex]\begin{gathered} u^{\prime}=(1)x^{1-1}+0 \\ u^{\prime}=(1)x^0 \\ u^{\prime}=(1)(1) \\ u^{\prime}=1 \end{gathered}[/tex][tex]\begin{gathered} v^{\prime}=2(2)x^{2-1}+2(1)x^{1-1}+0 \\ v^{\prime}=4x^1+2x^0 \\ v^{\prime}=4x+2 \end{gathered}[/tex]
Substitute them in the rule above
[tex]f^{\prime}(x)=\frac{1(2x^2+2x+3)-(x+1)(4x+2)}{(2x^2+2x+3)^2}[/tex]
Simplify the numerator
[tex]\begin{gathered} f^{\prime}(x)=\frac{2x^2+2x+3-\lbrack x(4x)+x(2)+1(4x)+1(2)\rbrack}{(2x^2+2x+3)^2} \\ f^{\prime}(x)=\frac{2x^2+2x+3-\lbrack4x^2+2x+4x+2\rbrack}{(2x^2+2x+3)^2} \\ f^{\prime}(x)=\frac{2x^2+2x+3-4x^2-2x-4x-2}{(2x^2+2x+3)^2} \end{gathered}[/tex]
Add the like terms in the denominator
[tex]\begin{gathered} f^{\prime}(x)=\frac{(2x^2-4x^2)+(2x-2x-4x)+(3-2)}{(2x^2+2x+3)} \\ f^{\prime}(x)=\frac{-2x^2-4x+1}{(2x^2+2x+3)^2} \end{gathered}[/tex]
