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

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Answer 1 · 2017-02-27T17:21:44

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

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

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#