How do you differentiate f(x)= (x^2-5x+2)/ (6x+1 )f(x)=x2−5x+26x+1 using the quotient rule?

1 Answer
Jarob R G.
May 3, 2018

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.

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