This question: $4$

point$($s$)$ possible

Su

Use the quotient rule to find the derivative.

$f(x)=\frac{(4x-1)(2x^{2}3)}{4x3}$

A$.f^{\text{'}}(x)=\frac{64x^{3}72x^2-12x48}{(4x3{)}^2}$

B$.f^{\text{'}}(x)=\frac{32x^{3}64x^{2}12x48}{(4x3{)}^2}$

C$.f^{\text{'}}(x)=\frac{64x^{3}64x^2-12x48}{(4x3{)}^2}$

D$.f^{\text{'}}(x)=\frac{64x^{3}64x^2-12x48}{4x3}$

Time Remaining: $01$:$02$:$38$