Answer: This question can be solved using the Quotient rule, which is defined as follows:
[tex]\begin{gathered} f(x)=\frac{g(x)}{h(x)} \\ f^{\prime}(x)=\frac{g^{\prime}(x)h(x)-g(x)h^{\prime}(x)}{h(x)^2} \end{gathered}[/tex]
For the given function:
[tex]\begin{gathered} f(x)=\frac{(x+1)}{2x^2+2x+3}\Rightarrow(1) \\ g(x)=(x+1) \\ h(x)=(2x^2+2x+3) \end{gathered}[/tex]
Therefore the derivative of (1) is:
[tex]f^{\prime}(x)=\frac{(x+1)^{\prime}(2x^2+2x+3)+(x+1)(2x^2+2x+3)^{\prime}}{(2x^2+2x+3)\cdot(2x^2+2x+3)}\Rightarrow(2)[/tex]
Simplifying the (2) gives the following result:
[tex]\begin{gathered} f^{\prime}(x)=\frac{(2x^2+2x+3)+(4x^2+10x+4)^{}}{(2x^2+2x+3)\cdot(2x^2+2x+3)} \\ f^{\prime}(x)=\frac{(6x^2+12x+7)^{}}{(2x^2+2x+3)\cdot(2x^2+2x+3)} \\ f^{\prime}(x)=\frac{(6x^2+12x+7)^{}}{(4x^4+8x^3+16x^2+12x+9)} \end{gathered}[/tex]
