Differentiate with respect to x using the quotient rule?

Y= (2x^{4}– 3x) / (4x -1)

d/dx [2x^{4}– 3x] = 8x^{3}– 3

and

d/dx [4x – ] = 4

by the quotient rule

f(x) = (g(x)) / (h(x)) , f’(x) = [g^{1}(x) h(x) – g(x)h^{1}(x)] / [h(x)]^{2 }

choosing g(x) = 2x^{4}– 3x and h(x) = 4x - 1

gives

dy/dx = [(8x^{3}– 3) (4x – 1) - (2x^{4}– 3x) (4) ] / (4x – 1)^{ 2}

therefore

dy/dx = (23x^{4}– 8x^{3}– 12x + 3 – 8x^{4}+ 12x) / 16x^{2}– 8x + 1

therefore

dy/dx = (24x^{4}– 8x^{3}+ 3) / (16x^{2}– 8x + 1)