Question

How do you differentiate `f(x)= (x^2-5x+2)/ (6x+1 )` using the quotient rule?

Answer 1

`f'(x) = (6x^2 + 2x - 17)/(6x+1)^2`

Explanation:

I've always remembered this rule by the saying "Low d-high minus high d-low, all over the square of what's below." This translates into the following procedure:

Suppose you have some function `f(x) = (g(x))/(h(x))`, where `g(x)` and `h(x)` are some functions of `x`. Then the derivative of `f(x)` is `f'(x) = (h(x)g'(x) - g(x)h'(x))/(h^2(x))`.

Solving the problem you presented, we have:

`f'(x) = ((6x+1) * d/(dx) (x^2 - 5x + 2) - (x^2 - 5x + 2) * d/(dx) (6x+1)) / (6x+1)^2`
`= ((6x+1)(2x-5) - (x^2-5x+2)(6))/(6x+1)^2`
`= ((12x^2 + 2x - 30x - 5) - (6x^2 - 30x + 12))/(6x+1)^2`
`= (12x^2 - 28x - 5 - 6x^2 + 30x - 12)/(6x+1)^2`
`= (6x^2 + 2x - 17)/(6x+1)^2 = f'(x)`

This is the final answer.