Question

Please can you help me find the derivative of the equation

Answer 1

The given function is

`f(x)=(x+1)/(2x^2+2x+3)`

To find its derivative we should use the rule of division

`(d)/(dx)((u)/(v))=(u^(\prime)v-uv^(\prime))/(v^2)`

Let

`\begin{gathered} u=x+1 \\ v=2x^2+2x+3 \end{gathered}`

Let us find u' and v' first

`\begin{gathered} u^(\prime)=(1)x^(1-1)+0 \\ u^(\prime)=(1)x^0 \\ u^(\prime)=(1)(1) \\ u^(\prime)=1 \end{gathered}`
`\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}`

Substitute them in the rule above

`f^(\prime)(x)=(1(2x^2+2x+3)-(x+1)(4x+2))/((2x^2+2x+3)^2)`

Simplify the numerator

`\begin{gathered} f^(\prime)(x)=(2x^2+2x+3-\lbrack x(4x)+x(2)+1(4x)+1(2)\rbrack)/((2x^2+2x+3)^2) \\ f^(\prime)(x)=(2x^2+2x+3-\lbrack4x^2+2x+4x+2\rbrack)/((2x^2+2x+3)^2) \\ f^(\prime)(x)=(2x^2+2x+3-4x^2-2x-4x-2)/((2x^2+2x+3)^2) \end{gathered}`

Add the like terms in the denominator

`\begin{gathered} f^(\prime)(x)=((2x^2-4x^2)+(2x-2x-4x)+(3-2))/((2x^2+2x+3)) \\ f^(\prime)(x)=(-2x^2-4x+1)/((2x^2+2x+3)^2) \end{gathered}`