Question

How do you find the first and second derivatives of `y= (6x^2+4)/ (2x+2)` using the quotient rule?

Answer 1

`y'=(2x^2+6x-2)/(x+1)^2`
`y''=10/(x+1)^3`

Explanation:

Using the Quotient rule we get

`y'=(6x(x+1)-(3x^2+2))/(x+1)^2`
which simplifyes to

`y'=(3x^2+6x-2)/(x+1)^2`

Again by the Quotient rule we get

`y''=((6x+6)(x+1)^2-(3x^2+6x-2)(2(x+1)))/(x+1)^3`
Simplifying we obtain
`y''=10/(x+1)^3`