Question

Please can you help me the derivative of the equation

Answer 1

Answer: This question can be solved using the Quotient rule, which is defined as follows:

`\begin{gathered} f(x)=(g(x))/(h(x)) \\ f^(\prime)(x)=(g^(\prime)(x)h(x)-g(x)h^(\prime)(x))/(h(x)^2) \end{gathered}`

For the given function:

`\begin{gathered} f(x)=((x+1))/(2x^2+2x+3)\Rightarrow(1) \\ g(x)=(x+1) \\ h(x)=(2x^2+2x+3) \end{gathered}`

Therefore the derivative of (1) is:

`f^(\prime)(x)=((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)`

Simplifying the (2) gives the following result:

`\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}`