What is the first derivative of y = (4x^2-6x+9)/(2x-1)

Quotient rule: f(x) = u/v, f'(x) = (vu' - uv')/(v^2)

(2x - 1)(8x - 6) - (4x^2 - 6x + 9)(2)
--------------------------------------…
(2x - 1)^2

16x^2 - 20x + 6 - 8x^2 + 12x - 18
--------------------------------------…
(2x - 1)^2

8x^2 - 8x - 12
----------------------
(2x - 1)^2

4(2x^2 - 2x - 3)
-------------------------- <===ANSWER
(2x - 1)^2

For division, the formula for derivatives is [(bottom)(derivative of top) - (top)(derivative of bottom)]/ (bottom squared)
y'= [(2x-1)(8x-6) - (4x^2 -6x+9)(2)]/ (2x-1)^2
Reduced:
y'= (8x^2 - 32x + 24) / (4x^2 - 4x +1)
Depending on how your teacher wants you to write your final answer, you may not want to reduce.

Use the quotient rule of differentiation: d/dx[u/v] = (vu ' - u v') / (v^(2))

u = 4x^(2) - 6x + 9
u ' = 8x - 6
v = 2x - 1
v ' = 2
v^(2) = (2x - 1)^(2)

y(x) = (4x^(2) - 6x + 9) / (2x - 1)

Take the derivative to both sides of the function with respect to x.

d/dx[y(x)] = d/dx[ (4x^(2) - 6x + 9) / (2x - 1) ]

y ' (x) = [(2x - 1) * (8x - 6) - (4x^(2) - 6x + 9) * (2) ] / [(2x - 1)^(2)]

Multiply using the distributive property to the numerator.

y ' (x) = [16x^(2) - 12x - 8x + 6 - 8x^(2) + 12x - 18] / [ (2x - 1)^(2)]

Combine like terms in the numerator.

y ' (x) = [8x^(2) - 8x - 12] / [ (2x - 1)^(2) ]

Factor out a 4 from the numerator.

y ' (x) = [ 4 [2x^(2) - 2x - 3] ] / [(2x - 1)^(2) ] ----> ANSWER