How do you differentiate y=(4x^2-3x+1)/(5x-3)?

1 Answer
Feb 27, 2017

Use the quotient rule: (u/v)' = (vu' - uv')/v^2

f(x)' =(4(5x^2-6x+1))/(5x-3)^2

Explanation:

Use the quotient rule: (u/v)' = (vu' - uv')/v^2

Let u = 4x^2-3x+1, so u' = 8x-3

Let v = 5x-3, so v' = 5

f(x)' = ((5x-3)(8x-3)-(4x^2-3x+1)(5))/(5x-3)^2

Distribute the numerator:

f(x)' = (40x^2-15x-24x+9-20x^2+15x-5)/(5x-3)^2

Simplify numerator:

f(x)' = (20x^2-24x+4)/(5x-3)^2 = (4(5x^2-6x+1))/(5x-3)^2